GRAVITATION-Exercise

Q1. Answer the following :
(a)You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
(b)An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why ?

Solution

(a) Electrical shielding is possible because electric charges can rearrange themselves on the surface of a conductor in such a way that the electric field inside the hollow region becomes zero. Gravity, however, is fundamentally different in nature. There is no concept of negative mass that could rearrange itself to cancel gravitational fields. Since all masses attract each other and gravitational interaction cannot be screened or neutralized, placing a body inside a hollow sphere or using any other arrangement cannot shield it from gravitational influence. The body will still experience the gravitational effects of nearby matter.

(b) An astronaut inside a small spacecraft orbiting the Earth feels weightless not because gravity is absent, but because both the astronaut and the spacecraft are falling freely under Earth’s gravity with the same acceleration. This situation remains unchanged even if the space station is very large. As long as the astronaut and the station are in the same orbit and share the same free-fall motion, no gravitational force is felt internally. Therefore, increasing the size of the space station does not help in detecting gravity through direct mechanical effects.

(c) The gravitational force exerted by a celestial body depends on its mass and distance, and it is indeed true that the Sun exerts a larger gravitational force on the Earth than the Moon does. However, tidal effects arise not from the total gravitational force, but from the variation of gravitational force across the Earth. This variation, called the tidal force, depends on how rapidly gravity changes with distance.

Using the idea of tidal force, we can express its dependence as

\[ \begin{aligned} \text{Tidal force} &\propto \frac{M}{r^{3}} \end{aligned} \]

Although the Sun has a much larger mass, it is also very far from the Earth. The Moon, despite being much less massive, is much closer to the Earth. Because the tidal force varies inversely as the cube of the distance, the Moon produces a larger difference in gravitational pull between the near and far sides of the Earth. As a result, the Moon’s tidal effect on the Earth is stronger than that of the Sun.

Q2. Choose the correct alternative :
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula \(–GMm(1/r_2 – 1/r_1)\) is more/less accurate than the formula \(mg(r_2 – r_1)\) for the difference of potential energy between two points \(r_2\) and \(r_1\) distance away from the centre of the earth.

Solution

(a) The acceleration due to gravity decreases with increasing altitude. As we move away from the surface of the Earth, the distance from the Earth’s centre increases, reducing the gravitational pull. This dependence can be written as

\[ \begin{aligned} g &= \frac{GM}{(R+h)^2} \end{aligned} \]

where \(R\) is the radius of the Earth and \(h\) is the altitude above its surface. Since \((R+h)\) increases with altitude, the value of \(g\) becomes smaller.

(b) The acceleration due to gravity decreases with increasing depth below the Earth’s surface, assuming the Earth to be a sphere of uniform density. As one goes deeper, only the mass enclosed within that radius contributes to gravity. The variation can be expressed as

\[ \begin{aligned} g_d &= g\left(1 - \frac{d}{R}\right) \end{aligned} \]

where \(d\) is the depth below the surface and \(R\) is the Earth’s radius. This shows that \(g\) decreases linearly with depth and becomes zero at the centre.

(c) The acceleration due to gravity is independent of the mass of the body experiencing the force. From Newton’s law of gravitation and the definition of acceleration, we have

\[ \begin{aligned} F &= \frac{GMm}{r^2}, \\ a &= \frac{F}{m} = \frac{GM}{r^2} \end{aligned} \]

The mass \(m\) of the body cancels out, showing that the acceleration depends only on the mass of the Earth and the distance from its centre, not on the mass of the falling body.

(d) The formula \(-GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)\) is more accurate than the expression \(mg(r_2 - r_1)\) for the change in gravitational potential energy between two points. The exact expression is derived directly from Newton’s law of gravitation and is valid for all values of \(r_1\) and \(r_2\).

The simpler expression \(mg(r_2 - r_1)\) assumes that the acceleration due to gravity remains constant, which is only a good approximation when the height difference \((r_2 - r_1)\) is very small compared to the Earth’s radius. Hence, the first formula provides a more accurate result in general.

Q3. Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?

Solution

If a planet goes around the Sun twice as fast as the Earth, its time period of revolution is half that of the Earth. Let the Earth’s orbital radius be \(R\) with time period \(T\), and let the orbital radius and time period of the planet be \(r\) and \(T/2\) respectively.

According to Kepler’s third law, the square of the time period of revolution is proportional to the cube of the orbital radius. This relation can be written as

\[ \begin{aligned} T^2 &\propto R^3, \\ \left(\frac{T}{2}\right)^2 &\propto r^3 \end{aligned} \]

Dividing the second relation by the first, we obtain

\[ \begin{aligned} \frac{(T/2)^2}{T^2} &= \frac{r^3}{R^3}, \\ \frac{1}{4} &= \frac{r^3}{R^3} \end{aligned} \]

Taking the cube root of both sides gives

\[ \begin{aligned} \frac{r}{R} &= \left(\frac{1}{4}\right)^{1/3} = \left(\frac{1}{2}\right)^{2/3} \end{aligned} \]

Hence, the orbital size of the planet is smaller than that of the Earth. Numerically, \(\left(\tfrac{1}{2}\right)^{2/3} \approx 0.63\), which means the planet’s orbit has a radius about \(0.63\) times the Earth’s orbital radius.

Q4. Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is \(\mathrm{4.22 × 10^8}\) m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Solution

Io revolves around Jupiter under the gravitational attraction of the planet. For a satellite moving in a nearly circular orbit, the mass of the central body can be determined by equating the gravitational force to the required centripetal force. This leads to the relation

\[ \begin{aligned} \frac{GM_J m}{r^2} &= \frac{m v^2}{r} \end{aligned} \]

Using \(v = \frac{2\pi r}{T}\), the above expression becomes

\[ \begin{aligned} M_J &= \frac{4\pi^2 r^3}{G T^2} \end{aligned} \]

For Io, the orbital radius is \(r = 4.22 \times 10^8 \, \text{m}\) and the orbital period is \(T = 1.769\) days. Converting the period into seconds,

\[ \begin{aligned} T &= 1.769 \times 24 \times 3600 \\ &\approx 1.53 \times 10^5 \, \text{s} \end{aligned} \]

Substituting the values into the expression for \(M_J\),

\[ \begin{aligned} M_J &= \frac{4\pi^2 (4.22 \times 10^8)^3}{(6.67 \times 10^{-11})(1.53 \times 10^5)^2} \\ &\approx 1.9 \times 10^{27} \, \text{kg} \end{aligned} \]

The mass of the Sun is approximately \(2.0 \times 10^{30} \, \text{kg}\). Therefore, the ratio of the mass of Jupiter to the mass of the Sun is

\[ \begin{aligned} \frac{M_J}{M_{\odot}} &= \frac{1.9 \times 10^{27}}{2.0 \times 10^{30}} \\ &\approx 9.5 \times 10^{-4} \end{aligned} \]

This shows that the mass of Jupiter is of the order of one-thousandth of the mass of the Sun, as required.

Q5. Let us assume that our galaxy consists of \(\mathrm{2.5 × 10^{11}}\) stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be \(\mathrm{10^5}\) ly.

Solution

We assume that the mass of the Milky Way is concentrated near its centre and is equal to the total mass of its stars. The galaxy contains \(2.5 \times 10^{11}\) stars, each of one solar mass. Taking the mass of the Sun as \(M_\odot = 2.0 \times 10^{30}\,\text{kg}\), the total mass of the galaxy is

\[ \begin{aligned} M &= 2.5 \times 10^{11} \times 2.0 \times 10^{30} \\ &= 5.0 \times 10^{41}\,\text{kg} \end{aligned} \]

The diameter of the Milky Way is \(10^5\) light years, so a star at a distance of \(50{,}000\) light years lies at a radius

\[ \begin{aligned} r &= 5.0 \times 10^{4} \times 9.46 \times 10^{15} \\ &\approx 4.73 \times 10^{20}\,\text{m} \end{aligned} \]

For circular motion under gravity, the time period of revolution is given by

\[ \begin{aligned} T &= 2\pi \sqrt{\frac{r^{3}}{GM}} \end{aligned} \]

Substituting \(G = 6.67 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}\), \(r = 4.73 \times 10^{20}\,\text{m}\), and \(M = 5.0 \times 10^{41}\,\text{kg}\),

\[ \begin{aligned} T &= 2\pi \sqrt{\frac{(4.73 \times 10^{20})^{3}}{6.67 \times 10^{-11} \times 5.0 \times 10^{41}}} \\ &\approx 3.5 \times 10^{16}\,\text{s} \end{aligned} \]

Converting this time into years,

\[ \begin{aligned} T &= \frac{3.5 \times 10^{16}}{3.15 \times 10^{7}} \\ &\approx 1.1 \times 10^{9}\,\text{years} \end{aligned} \]

Thus, under the given assumptions, a star located about \(50{,}000\) light years from the centre of the Milky Way would take of the order of a billion years to complete one revolution around the galactic centre.

Q6. Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.

Solution

(a) When the zero of gravitational potential energy is taken at infinity, the potential energy of an orbiting satellite at a finite distance from the Earth is negative. For a satellite in a circular orbit, the gravitational attraction provides the required centripetal force. This leads to the relations

\[ \begin{aligned} \frac{GMm}{r^2} &= \frac{m v^2}{r}, \\ v^2 &= \frac{GM}{r} \end{aligned} \]

The kinetic energy of the satellite is therefore

\[ \begin{aligned} K &= \frac{1}{2} m v^2 = \frac{GMm}{2r} \end{aligned} \]

The gravitational potential energy at distance \(r\) is

\[ \begin{aligned} U &= -\frac{GMm}{r} \end{aligned} \]

Hence, the total mechanical energy of the satellite is

\[ \begin{aligned} E &= K + U \\\\ &= \frac{GMm}{2r} - \frac{GMm}{r} \\\\ &= -\frac{GMm}{2r} \end{aligned} \]

This shows that the total energy of an orbiting satellite is negative and is equal to the negative of its kinetic energy.

(b) An orbiting satellite already possesses kinetic energy due to its orbital motion. To remove it completely from the Earth’s gravitational influence, its total mechanical energy must be raised from a negative value to zero. The additional energy required is therefore equal in magnitude to its existing total energy.

A stationary object at the same height has zero kinetic energy and a negative potential energy of the same magnitude as that of the satellite. To send this object out of Earth’s gravitational field, it must be given enough energy to raise its total energy from this negative value to zero, which requires a larger energy input than in the case of the orbiting satellite.

Hence, the energy required to launch an orbiting satellite out of Earth’s gravitational influence is less than the energy required to project a stationary object from the same height out of Earth’s influence.

Q7. Does the escape speed of a body from the earth depend on
(a) the mass of the body,
(b) the location from where it is projected,
(c) the direction of projection, (d) the height of the location from where the body is launched?

Solution

The escape speed of a body is the minimum speed required for it to move away from the Earth and reach infinity with zero residual speed. This condition is obtained by equating the initial kinetic energy of the body to the gravitational potential energy needed to overcome the Earth’s attraction.

If a body of mass \(m\) is projected from the Earth’s surface of radius \(R\), the energy balance can be written as

\[ \begin{aligned} \frac{1}{2} m v_e^2 &= \frac{GMm}{R} \end{aligned} \]

Solving for the escape speed \(v_e\),

\[ \begin{aligned} v_e &= \sqrt{\frac{2GM}{R}} \end{aligned} \]

From this expression, it is clear that the escape speed does not depend on the mass of the body, since the mass \(m\) cancels out during the derivation. Hence, bodies of different masses require the same escape speed when launched from the same location.

The escape speed also does not depend on the direction of projection. Only the magnitude of the velocity matters, because gravitational force is central and acts along the line joining the centres of the Earth and the body.

However, the escape speed does depend on the location and the height from where the body is projected. If the body is launched from a height \(h\) above the Earth’s surface, the required escape speed becomes

\[ \begin{aligned} v_e &= \sqrt{\frac{2GM}{R+h}} \end{aligned} \]

Since \(R+h\) is larger than \(R\), the escape speed decreases with increasing height. Therefore, the escape speed is independent of the mass of the body and the direction of projection, but it does depend on the location and the height from which the body is launched.

Q8. A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant
(a) linear speed,
(b) angular speed, ,
(c) angular momentum,
(d) kinetic energy,
(e) potential energy,
(f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

Solution

A comet moving around the Sun in a highly elliptical orbit is governed by the central gravitational force of the Sun. Since this force always acts along the line joining the comet and the Sun, it produces no torque about the Sun. As a result, certain physical quantities remain conserved while others change continuously along the orbit.

(a) The linear speed of the comet is not constant. When the comet is closer to the Sun, the gravitational attraction is stronger and the comet moves faster. When it is far from the Sun, the attraction weakens and the comet slows down. This variation in speed is a direct consequence of energy conservation and the changing distance from the Sun.

(b) The angular speed of the comet is also not constant. As the comet approaches the Sun, it sweeps out angular displacement more rapidly, while far from the Sun it sweeps out angles more slowly. This behaviour is consistent with Kepler’s second law, which states that equal areas are swept out in equal intervals of time.

(c) The angular momentum of the comet about the Sun remains constant throughout its motion. Since the gravitational force is central, the torque about the Sun is zero, and hence angular momentum is conserved despite changes in speed and direction.

(d) The kinetic energy of the comet is not constant. It increases as the comet moves closer to the Sun and gains speed, and decreases as the comet moves away and slows down. This change in kinetic energy is balanced by a corresponding change in gravitational potential energy.

(e) The gravitational potential energy of the comet is also not constant. It becomes more negative when the comet is closer to the Sun and less negative when the comet is farther away. Thus, potential energy varies continuously along the elliptical orbit.

(f) However, the total mechanical energy of the comet remains constant throughout the orbit. In the absence of any non-conservative forces and neglecting mass loss, the sum of kinetic and potential energies stays fixed and negative, which is characteristic of a bound elliptical orbit.

Q9. Which of the following symptoms is likely to afflict an astronaut in space
(a) swollen feet,
(b) swollen face,
(c) headache,
(d) orientational problem.

Solution

An astronaut in space experiences a condition of near weightlessness because both the astronaut and the spacecraft are in continuous free fall around the Earth. In the absence of the normal gravitational pull that causes fluids in the body to collect in the lower limbs, several physiological changes take place.

Swollen feet are not likely to occur in space. On Earth, gravity causes blood and other body fluids to accumulate in the legs and feet, but in a weightless environment this downward pooling does not happen.

A swollen face is a common symptom experienced by astronauts. In microgravity, body fluids redistribute more uniformly and tend to shift toward the upper part of the body, leading to facial puffiness often described as a “puffy face” appearance.

Headaches may occur during the initial period of spaceflight due to fluid shifts, changes in blood pressure regulation, and adaptation to the new environment. Thus, headaches are a likely symptom, especially during the early stages of a mission.

Orientational problems are also likely to affect astronauts. In the absence of gravity, the inner ear and other balance-related sensory systems do not receive their usual cues, which can lead to disorientation and difficulty in distinguishing up from down.

Therefore, among the given options, swollen face, headache, and orientational problems are likely to afflict an astronaut in space, whereas swollen feet are not.

Q11. For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Solution

At an arbitrary point \(P\) inside the hemispherical shell, the gravitational intensity is again obtained by vectorially adding the contributions from all mass elements of the shell. Unlike the centre, the point \(P\) does not possess complete symmetry with respect to the mass distribution.

Because the shell lies entirely below the grey flat surface, the vertical components of gravitational intensity due to all mass elements still add up in the downward direction. Hence, the resultant field at point \(P\) must have a downward component.

At the same time, since point \(P\) is displaced to the left of the axis of symmetry, the horizontal components of the gravitational field no longer cancel completely. The mass distribution is effectively heavier on the right side of point \(P\), so the net horizontal component of gravitational intensity points toward the right, that is, toward the interior and axis of the hemisphere.

Combining these two effects, the resultant gravitational intensity at point \(P\) is directed downward and toward the right. This corresponds to the direction shown by arrow f.

Q12. A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun \(\mathrm{= 2×10^{30}\ kg}\), mass of the earth \(\mathrm{= 6×10^{24}\ kg}\). Neglect the effect of other planets etc. (orbital radius \(\mathrm{= 1.5 × 10^{11}\ m}\)).

Solution

As the rocket travels from the Earth toward the Sun, it is attracted by the gravitational forces of both bodies. At a certain point along the line joining their centres, these two gravitational forces act in opposite directions and can balance each other. At this point, the net gravitational force on the rocket becomes zero.

Let the distance between the centres of the Earth and the Sun be \(R = 1.5 \times 10^{11}\,\text{m}\). Suppose the rocket is at a distance \(x\) from the Earth’s centre along the line toward the Sun. Then its distance from the Sun’s centre is \((R - x)\).

The gravitational force exerted by the Earth on the rocket is

\[ \begin{aligned} F_E &= \frac{G M_E m}{x^2} \end{aligned} \]

and the gravitational force exerted by the Sun on the rocket is

\[ \begin{aligned} F_S &= \frac{G M_S m}{(R - x)^2} \end{aligned} \]

At the point where the net gravitational force is zero, these two forces are equal in magnitude:

\[ \begin{aligned} \frac{G M_E m}{x^2} &= \frac{G M_S m}{(R - x)^2} \end{aligned} \]

Canceling the common factors and substituting \(M_E = 6 \times 10^{24}\,\text{kg}\) and \(M_S = 2 \times 10^{30}\,\text{kg}\),

\[ \begin{aligned} \frac{6 \times 10^{24}}{x^2} &= \frac{2 \times 10^{30}}{(R - x)^2} \end{aligned} \]

Rearranging and simplifying,

\[ \begin{aligned} \frac{R - x}{x} &= \sqrt{\frac{2 \times 10^{30}}{6 \times 10^{24}}}\\\\ &= \sqrt{\frac{1}{3} \times 10^{6}}\\\\ &\approx 577 \end{aligned} \]

This gives

\[ \begin{aligned} R - x &= 577x \\ R &= 578x \\ x &= \frac{R}{578} \end{aligned} \]

Substituting \(R = 1.5 \times 10^{11}\,\text{m}\),

\[ \begin{aligned} x &\approx \frac{1.5 \times 10^{11}}{578}\\\\ &\approx 2.6 \times 10^{8}\,\text{m} \end{aligned} \]

Hence, the gravitational force on the rocket becomes zero at a distance of approximately \(2.6 \times 10^{8}\,\text{m}\) from the Earth’s centre, along the line joining the Earth and the Sun.

Q13. How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is \(1.5 × 10^8\) km.

Solution

The mass of the Sun can be estimated by studying the motion of the Earth around it. The Earth moves in a nearly circular orbit under the gravitational attraction of the Sun, and this gravitational force provides the necessary centripetal force for the Earth’s orbital motion.

Let the mass of the Sun be \(M_\odot\), the mass of the Earth be \(m\), the mean orbital radius be \(r = 1.5 \times 10^{8}\,\text{km} = 1.5 \times 10^{11}\,\text{m}\), and the time period of revolution of the Earth be \(T = 1\) year \(= 3.15 \times 10^{7}\,\text{s}\).

The gravitational force exerted by the Sun on the Earth is

\[ \begin{aligned} F &= \frac{G M_\odot m}{r^2} \end{aligned} \]

The centripetal force required to keep the Earth in its circular orbit is

\[ \begin{aligned} F &= \frac{m v^2}{r} = \frac{m}{r}\left(\frac{2\pi r}{T}\right)^2 \end{aligned} \]

Equating the gravitational force to the centripetal force,

\[ \begin{aligned} \frac{G M_\odot m}{r^2} &= \frac{m}{r}\left(\frac{2\pi r}{T}\right)^2 \end{aligned} \]

Canceling the mass of the Earth and simplifying,

\[ \begin{aligned} M_\odot &= \frac{4\pi^2 r^3}{G T^2} \end{aligned} \]

Substituting \(r = 1.5 \times 10^{11}\,\text{m}\), \(T = 3.15 \times 10^{7}\,\text{s}\), and \(G = 6.67 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}\),

\[ \begin{aligned} M_\odot &\approx \frac{4\pi^2 (1.5 \times 10^{11})^3}{6.67 \times 10^{-11} (3.15 \times 10^{7})^2} \\ &\approx 2.0 \times 10^{30}\,\text{kg} \end{aligned} \]

Thus, by analysing the Earth’s orbital motion, the mass of the Sun can be estimated to be approximately \(2 \times 10^{30}\,\text{kg}\).

Q14. A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is \(\mathrm{1.50 × 10^8\ km}\) away from the sun ?

Solution

The motion of planets around the Sun follows Kepler’s third law, which states that the square of the time period of revolution of a planet is proportional to the cube of the mean distance of the planet from the Sun.

Let the Earth’s orbital radius be \(R_E = 1.50 \times 10^8 \,\text{km}\) and its time period be \(T_E\). For Saturn, let the orbital radius be \(R_S\) and the time period be \(T_S = 29.5\,T_E\).

According to Kepler’s third law,

\[ \begin{aligned} \frac{T_S^2}{T_E^2} &= \frac{R_S^3}{R_E^3} \end{aligned} \]

Substituting \(T_S = 29.5\,T_E\),

\[ \begin{aligned} (29.5)^2 &= \frac{R_S^3}{R_E^3} \end{aligned} \]

Taking the cube root of both sides,

\[ \begin{aligned} \frac{R_S}{R_E} &= (29.5)^{2/3} \end{aligned} \]

Evaluating the numerical value,

\[ \begin{aligned} (29.5)^{2/3} &\approx 9.6 \end{aligned} \]

Hence, the distance of Saturn from the Sun is

\[ \begin{aligned} R_S &= 9.6 \times 1.50 \times 10^8 \\ &\approx 1.44 \times 10^9 \,\text{km} \end{aligned} \]

Therefore, Saturn is approximately \(1.44 \times 10^9\) km away from the Sun.

Q15. A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ?

Solution

The weight of a body on the surface of the Earth is equal to the gravitational force acting on it at the Earth’s surface. If the body weighs \(63\,\text{N}\), then this force can be written as

\[ \begin{aligned} F_0 &= \frac{GMm}{R^2} = 63\,\text{N} \end{aligned} \]

where \(R\) is the radius of the Earth. When the body is raised to a height equal to half the Earth’s radius, the distance from the Earth’s centre becomes

\[ \begin{aligned} r &= R + \frac{R}{2} = \frac{3R}{2} \end{aligned} \]

The gravitational force on the body at this height is then

\[ \begin{aligned} F &= \frac{GMm}{r^2} = \frac{GMm}{\left(\frac{3R}{2}\right)^2} \end{aligned} \]

Dividing this expression by the force at the Earth’s surface,

\[ \begin{aligned} \frac{F}{F_0} &= \frac{R^2}{\left(\frac{3R}{2}\right)^2} = \frac{4}{9} \end{aligned} \]

Hence, the gravitational force on the body at the given height is

\[ \begin{aligned} F &= \frac{4}{9} \times 63 = 28\,\text{N} \end{aligned} \]

Therefore, the gravitational force acting on the body at a height equal to half the Earth’s radius is \(28\,\text{N}\).

Q16. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ?

Solution

When the Earth is assumed to be a sphere of uniform mass density, the acceleration due to gravity inside the Earth decreases linearly with depth. At a depth \(d\) below the surface, the acceleration due to gravity is given by

\[ \begin{aligned} g' &= g\left(1 - \frac{d}{R}\right) \end{aligned} \]

Here, \(g\) is the acceleration due to gravity at the surface and \(R\) is the radius of the Earth. The weight of the body on the surface is \(250\,\text{N}\), which means

\[ \begin{aligned} W &= mg = 250\,\text{N} \end{aligned} \]

Half way down to the centre of the Earth, the depth is \(d = \frac{R}{2}\). Substituting this value into the expression for \(g'\),

\[ \begin{aligned} g' &= g\left(1 - \frac{1}{2}\right) = \frac{g}{2} \end{aligned} \]

The weight of the body at this depth is

\[ \begin{aligned} W' &= mg' = m\left(\frac{g}{2}\right) = \frac{1}{2}mg \end{aligned} \]

Using \(mg = 250\,\text{N}\),

\[ \begin{aligned} W' &= \frac{1}{2} \times 250 = 125\,\text{N} \end{aligned} \]

Therefore, using the relation \(g' = g(1 - d/R)\), the body would weigh \(125\,\text{N}\) at a point half way down to the centre of the Earth.

Q17. A rocket is fired vertically with a speed of \(\mathrm{5\ km\ s^{-1}}\) from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = \(\mathrm{6.0 × 10^24\ kg}\); mean radius of the earth = \(\mathrm{6.4 × 10^6\ m;\ G = 6.67 × 10^{–11}\ N\ m^2\ kg^{–2}}\).

Solution

The rocket is fired vertically upward from the Earth’s surface with an initial speed \(v_0 = 5 \times 10^{3}\,\text{m s}^{-1}\). As it rises, its speed decreases under the influence of Earth’s gravity and becomes zero at the highest point. The maximum distance reached can be determined using the conservation of mechanical energy.

Let the mass of the rocket be \(m\), the mass of the Earth be \(M = 6.0 \times 10^{24}\,\text{kg}\), and the mean radius of the Earth be \(R = 6.4 \times 10^{6}\,\text{m}\). If the rocket reaches a maximum distance \(r\) from the Earth’s centre, its velocity at that point is zero.

At the Earth’s surface, the total mechanical energy of the rocket is

\[ \begin{aligned} E_{\text{initial}} &= \frac{1}{2} m v_0^2 - \frac{GMm}{R} \end{aligned} \]

At the highest point of ascent, the kinetic energy is zero and the total energy is purely gravitational potential energy,

\[ \begin{aligned} E_{\text{final}} &= -\frac{GMm}{r} \end{aligned} \]

Applying conservation of energy,

\[ \begin{aligned} \frac{1}{2} m v_0^2 - \frac{GMm}{R} &= -\frac{GMm}{r} \end{aligned} \]

Canceling the mass \(m\) and rearranging the terms,

\[ \begin{aligned} \frac{1}{r} &= \frac{1}{R} - \frac{v_0^2}{2GM} \end{aligned} \]

Substituting \(G = 6.67 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}\), \(M = 6.0 \times 10^{24}\,\text{kg}\), \(R = 6.4 \times 10^{6}\,\text{m}\), and \(v_0 = 5 \times 10^{3}\,\text{m s}^{-1}\),

\[ \begin{aligned} \frac{1}{r} &= \frac{1}{6.4 \times 10^{6}} - \frac{(5 \times 10^{3})^2}{2 \times 6.67 \times 10^{-11} \times 6.0 \times 10^{24}} \\ &\approx 1.56 \times 10^{-7} - 3.12 \times 10^{-8} \\ &\approx 1.25 \times 10^{-7} \end{aligned} \]

Taking the reciprocal,

\[ \begin{aligned} r &\approx 8.0 \times 10^{6}\,\text{m} \end{aligned} \]

Thus, the maximum distance of the rocket from the Earth’s centre is about \(8.0 \times 10^{6}\,\text{m}\). The height reached above the Earth’s surface is

\[ \begin{aligned} h &= r - R = 8.0 \times 10^{6} - 6.4 \times 10^{6} \\ &= 1.6 \times 10^{6}\,\text{m} \end{aligned} \]

Therefore, the rocket rises to a height of approximately \(1.6 \times 10^{6}\,\text{m}\), or about \(1600\,\text{km}\), above the Earth’s surface before returning to the Earth.

Q18. The escape speed of a projectile on the earth’s surface is \(\mathrm{11.2\ km\ s^{–1}}\). A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Solution

The escape speed from the Earth’s surface is the minimum speed required for a body to reach infinity with zero speed. It is given as \(v_e = 11.2\,\text{km s}^{-1}\). The body in this case is projected with a speed three times the escape speed, so its initial speed is

\[ \begin{aligned} v_0 &= 3v_e \end{aligned} \]

At the Earth’s surface, the total mechanical energy of the body is the sum of its kinetic energy and gravitational potential energy. Using the definition of escape speed, the potential energy at the surface can be written in terms of \(v_e\) as

\[ \begin{aligned} -\frac{GMm}{R} &= -\frac{1}{2} m v_e^2 \end{aligned} \]

Hence, the total energy at launch is

\[ \begin{aligned} E &= \frac{1}{2} m v_0^2 - \frac{GMm}{R} \\ &= \frac{1}{2} m (3v_e)^2 - \frac{1}{2} m v_e^2 \\ &= \frac{1}{2} m (9v_e^2 - v_e^2) \\ &= 4 m v_e^2 \end{aligned} \]

Far away from the Earth, the gravitational potential energy becomes zero. If the speed of the body at infinity is \(v_\infty\), the total energy there is purely kinetic,

\[ \begin{aligned} E &= \frac{1}{2} m v_\infty^2 \end{aligned} \]

Equating the total energies at the Earth’s surface and at infinity,

\[ \begin{aligned} \frac{1}{2} m v_\infty^2 &= 4 m v_e^2 \end{aligned} \]

Solving for \(v_\infty\),

\[ \begin{aligned} v_\infty &= \sqrt{8}\,v_e = 2\sqrt{2}\,v_e \end{aligned} \]

Substituting \(v_e = 11.2\,\text{km s}^{-1}\),

\[ \begin{aligned} v_\infty &= 2\sqrt{2} \times 11.2 \\ &\approx 31.7\,\text{km s}^{-1} \end{aligned} \]

Therefore, the speed of the body far away from the Earth is approximately \(31.7\,\text{km s}^{-1}\).

Q19. A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = \(\mathrm{6.0×10^{24}\ kg}\); radius of the earth = \(6.4 × 10^6\ m;\ G = 6.67 × 10^{–11}\ N\ m^2\ kg^{–2}\).

Solution

A satellite moving in a circular orbit around the Earth already possesses kinetic energy. To remove it completely from the Earth’s gravitational influence, its total mechanical energy must be raised from a negative value to zero. Hence, the energy required is equal in magnitude to the satellite’s total energy while in orbit.

Let the mass of the satellite be \(m = 200\,\text{kg}\), the mass of the Earth be \(M = 6.0 \times 10^{24}\,\text{kg}\), and the radius of the Earth be \(R = 6.4 \times 10^{6}\,\text{m}\). The satellite orbits at a height of \(400\,\text{km} = 4.0 \times 10^{5}\,\text{m}\), so the orbital radius is

\[ \begin{aligned} r &= R + h \\\\ &= 6.4 \times 10^{6} + 4.0 \times 10^{5} \\\\ &= 6.8 \times 10^{6}\,\text{m} \end{aligned} \]

For a satellite in a circular orbit, the total mechanical energy is given by

\[ \begin{aligned} E &= -\frac{GMm}{2r} \end{aligned} \]

The energy that must be supplied to take the satellite to infinity is therefore

\[ \begin{aligned} \Delta E &= \frac{GMm}{2r} \end{aligned} \]

Substituting \(G = 6.67 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}\),
\(M = 6.0 \times 10^{24}\,\text{kg}\),
\(m = 200\,\text{kg}\), and
\(r = 6.8 \times 10^{6}\,\text{m}\),

\[ \begin{aligned} \Delta E &= \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{2 \times 6.8 \times 10^{6}} \\\\ &\approx 5.9 \times 10^{9}\,\text{J} \end{aligned} \]

Thus, approximately \(5.9 \times 10^{9}\,\text{J}\) of energy must be expended to rocket the satellite out of the Earth’s gravitational influence.

Q20. Two stars each of one solar mass (= \(\mathrm{2×10^{30}\ kg}\)) are approaching each other for a head on collision. When they are a distance \(\mathrm{10^9\ km}\), their speeds are negligible. What is the speed with which they collide ? The radius of each star is \(\mathrm{10^4\ km}\). Assume the stars to remain undistorted until they collide. (Use the known value of G).

Solution

Two identical stars, each of mass \(M = 2.0 \times 10^{30}\,\text{kg}\), are initially very far apart and approach each other along a straight line under mutual gravitational attraction. When their separation is \(r_i = 10^{9}\,\text{km} = 1.0 \times 10^{12}\,\text{m}\), their speeds are negligible, so the initial kinetic energy may be taken as zero.

Each star has a radius of \(R = 10^{4}\,\text{km} = 1.0 \times 10^{7}\,\text{m}\). At the instant of collision, the distance between their centres is therefore

\[ \begin{aligned} r_f &= 2R = 2.0 \times 10^{7}\,\text{m} \end{aligned} \]

By conservation of mechanical energy, the decrease in gravitational potential energy is converted into kinetic energy of the two stars. The initial gravitational potential energy of the system is

\[ \begin{aligned} U_i &= -\frac{G M^2}{r_i} \end{aligned} \]

and the gravitational potential energy just before collision is

\[ \begin{aligned} U_f &= -\frac{G M^2}{r_f} \end{aligned} \]

If the speed of each star at collision is \(v\), the total kinetic energy of the system is

\[ \begin{aligned} K &= 2 \times \frac{1}{2} M v^2 = M v^2 \end{aligned} \]

Applying conservation of energy,

\[ \begin{aligned} M v^2 &= G M^2 \left( \frac{1}{r_f} - \frac{1}{r_i} \right) \end{aligned} \]

Canceling one factor of \(M\) and substituting \(G = 6.67 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}\),

\[ \begin{aligned} v^2 &= 6.67 \times 10^{-11} \times 2.0 \times 10^{30} \left( \frac{1}{2.0 \times 10^{7}} - \frac{1}{1.0 \times 10^{12}} \right) \\ &\approx 6.67 \times 10^{-11} \times 2.0 \times 10^{30} \times 5.0 \times 10^{-8} \\ &\approx 6.7 \times 10^{12} \end{aligned} \]

Taking the square root,

\[ \begin{aligned} v &\approx 2.6 \times 10^{6}\,\text{m s}^{-1} \end{aligned} \]

This is the speed of each star just before collision. Since the stars are moving toward each other with equal speeds, the speed of collision, that is, their relative speed at impact, is

\[ \begin{aligned} v_{\text{collision}} &= 2v \approx 5.2 \times 10^{6}\,\text{m s}^{-1} \end{aligned} \]

Thus, the two stars collide with a relative speed of about \(5.2 \times 10^{6}\,\text{m s}^{-1}\).

Q21. Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?

Solution

Two identical heavy spheres, each of mass \(M = 100\,\text{kg}\) and radius \(0.10\,\text{m}\), are placed with their centres \(1.0\,\text{m}\) apart. The midpoint of the line joining their centres lies at a distance of \(0.50\,\text{m}\) from the centre of each sphere. Since this point lies outside both spheres, each sphere may be treated as a point mass located at its centre.

The gravitational force at the midpoint due to one sphere is

\[ \begin{aligned} F_1 &= \frac{G M m}{(0.50)^2} \end{aligned} \]

where \(m\) is the mass of a test object placed at the midpoint. The second sphere produces a force of the same magnitude, but in the opposite direction, because the midpoint is symmetrically located between the two masses. Hence, the two gravitational forces cancel each other.

Therefore, the net gravitational force at the midpoint is zero.

The gravitational potential at a point due to a mass \(M\) at distance \(r\) is given by

\[ \begin{aligned} V &= -\frac{GM}{r} \end{aligned} \]

At the midpoint, the potential due to one sphere is

\[ \begin{aligned} V_1 &= -\frac{G \times 100}{0.50} \end{aligned} \]

The total gravitational potential at the midpoint, due to both spheres, is the algebraic sum of the two potentials:

\[ \begin{aligned} V &= -\frac{2G \times 100}{0.50}\\\\ &= -400G\\\\ &\approx -2.67 \times 10^{-8}\,\text{J kg}^{-1} \end{aligned} \]

Since the net gravitational force at the midpoint is zero, an object placed at this point is in equilibrium. However, this equilibrium is unstable. If the object is displaced slightly toward one sphere, the gravitational attraction toward that nearer sphere becomes stronger than the attraction due to the other sphere, causing the object to move further away from the midpoint instead of returning to it.

Thus, the gravitational force at the midpoint is zero, the gravitational potential there is finite and negative, and the equilibrium of an object placed at that point is unstable.