GRAVITATION-Exercise

Q1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Solution:
When the distance between two objects is reduced to half, the gravitational force between them increases by four times.
Let Initial Distance be R
New Distance =\(\frac{R}{2}\) \[ \begin{align} F_i=G\frac{m_1m_2}{R^2}\tag{1}\\\\ F_f=G\frac{m_1m_2}{(\frac{R}{2})^2}\tag{2}\\ \end{align} \] Dividing Eqn (1) by Eqn (2) \[ \begin{align} \frac{F_i}{F_f}&=\dfrac{G\dfrac{m_1m_2}{R^2}}{G\dfrac{m_1m_2}{(\frac{R}{2})^2}}\\\\ \frac{F_i}{F_f}&=\frac{\left(\frac{R}{2}\right)^2}{R^2}\\\\ \frac{F_i}{F_f}&=\frac{1}{4} \end{align} \] \[\boxed{F_f=4\times F_i}\] In short, reducing the distance to half makes the gravitational attraction four times stronger.

Q2. The gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

Solution:
Though the gravitational force acting on a heavy object is greater because of its larger mass, it also has more inertia, which resists changes in motion. According to Newton’s second law, acceleration is force divided by mass \(a=\frac{F}{m}\).
Since both force and mass increase proportionally for heavier objects, these effects cancel out, resulting in the same acceleration for both heavy and light objects.
Therefore, in the absence of air resistance, all objects fall at the same rate regardless of their mass.

Q3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the Earth is \(6 \times 10^{24}\) kg and radius of the earth is \(6.4 \times 10^6 m\))

Solution:
Mass of the Earth \((m_1)=6\times10^{24}\) kg
Mass of the Object \((m_2)=1\) kg
Distance between them = Radius of the of the Earth \(6.4\times10^6\)
G \(=6.67\times10{-11}Nm^2/kg^2\)
Gravitational Force= \[\small \begin{aligned} F&=G\frac{m_1m_2}{R^2}\\\\ &=6.67\times10^{-11}\times\frac{ 6\times10^{24}\times1}{(6.4\times10^6)^2}\\\\ &=\frac{6.67\times6}{6.4\times6.4}\times\frac{10^{-11}\times10^{24}}{10^{12}}\\\\ &=\frac{6.67\times6}{6.4\times6.4}\times{10^{(-11+24-12)}}\\\\ &=\frac{6.67\times6\times 10}{6.4\times6.4}\\\\ &=\frac{400.2}{40.96}\\\\ &\approx9.8\,N \end{aligned} \]

Q4. The Earth and the Moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

Solution:
The Earth and the Moon pull on each other with gravitational forces that are exactly equal in strength but opposite in direction.

This is because Newton’s third law tells us that every action has an equal and opposite reaction.

So, the force Earth applies to the Moon is the same size as the force the Moon applies to Earth. However, even though the forces are equal, their effects are very different.

The Moon is much smaller than the Earth, so it accelerates more and moves around the Earth, while the Earth’s huge mass barely shows any movement.

In short, both bodies feel the same strength of pull, but they respond differently because of their difference in mass.

Q5. If the Moon attracts the Earth, why does the Earth not move towards the Moon?

Solution:
The Earth does experience a pull from the Moon’s gravity, but it doesn't move noticeably towards the Moon because of its massive size compared to the Moon. Both the Earth and Moon pull on each other with equal force according to the law of gravitation.

However, since the Earth’s mass is much larger, this force causes only a tiny acceleration, so any movement towards the Moon is extremely slow and almost imperceptible.

The Moon, being smaller, moves noticeably around the Earth, while the Earth’s movement around the common center of mass is minimal and not easily seen.

This difference in mass means the Earth appears almost stationary as the Moon orbits it.

Q6. What happens to the force between two objects, if
(i) The mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) The masses of both objects are doubled?

Solution:

1. If the mass of one object is doubled, the gravitational force between the two objects also doubles because force is directly proportional to the product of their masses.
2. When the distance between the two objects is doubled, the force becomes one-fourth of the original due to the inverse square relationship. If the distance is tripled, the force decreases to one-ninth of the original force.
3. If both masses are doubled, the force increases by four times since the force depends on the product of the two masses.

Q7. What is the importance of the universal law of gravitation?

The universal law of gravitation is incredibly important because it helps us understand how every object in the universe is connected through the force of gravity.

It explains why planets orbit the Sun, why the Moon moves around the Earth, and how artificial satellites stay in orbit.

Beyond space, it also sheds light on natural phenomena we see on Earth, such as tides, rainfall, snowfall, and river flows.

In essence, this law lets us grasp how gravitational forces shape not only celestial movements but also influence life and environment on our planet.

It gives a unified explanation for forces acting at a distance, making it fundamental to physics and astronomy.

Q8. What is the acceleration of free fall?

Solution:
Acceleration of free fall is how much faster an object gets every second as it falls freely under Earth’s gravity. It’s a constant speed-up of about 9.8 meters per second squared near the Earth’s surface. This means every second, the falling object goes almost 10 m/s faster than the second before.

Q9. What do we call the gravitational force between the earth and an object?

Solution:
The gravitational force between the Earth and an object is called the weight of that object. This force is what pulls the object towards the Earth’s surface and gives it the sensation of heaviness.

Q10. Amit buys a few grams of gold at the poles as per the instructions of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of the gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

Solution:
Amit’s friend will not agree with the weight of gold when handed over at the equator because the weight of an object depends on gravity, which varies with location. Gravity is stronger at the poles and weaker at the equator due to Earth's shape and its rotation. So, even though the mass of gold remains the same, its weight measured at the equator will be less than that at the poles. This means the gold will weigh less at the equator, and the difference arises because the value of acceleration due to gravity (g) is greater at the poles than at the equator.

Q11. Why will a sheet of paper fall more slowly than one that is crumpled into a ball?

Solution:
A sheet of paper falls more slowly than a crumpled ball of paper because the flat sheet has a larger surface area that pushes against more air as it falls. This air resistance acts like a force holding it up and slows its fall. The crumpled paper, being compact with less surface area, faces less air resistance, so it drops faster through the air.

Q12. Gravitational force on the surface of the moon is only \(\frac{1}{6}\) as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on Earth?

Solution:
Mass of Obejct = 10 kg Weight \(W\) on the Earth = \[\begin{aligned}W&=mg\\&=10\times 9.8\\&=98N\end{aligned}\] Weight \(W\) on the Moon = \[\begin{aligned}W&=mg\\\\&=10\times \frac{9.8}{6}\\\\&\approx16.3N\end{aligned}\]

Q13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.

Solution:
(i) Initial Velocity \(u=49m/s\)
Final Velocity will be \(v=0\)
Acceleration due to gravity \(g=-9.8\) (acting against the motion)
Height \(h\) = \[\small \begin{aligned} v^2&=u^2+2gh\\ 0&=49^2+2\times (-9.8)\times h\\ -49^2&=-2\times 9.8\times h\\ \implies 49^2&=2\times 9.8\times h\\\\ \implies h&=\frac{49\times 49\times10}{2\times 98}\\\\ &=122.5\,m \end{aligned} \] (ii) Initial Velocity \(u=49m/s\)
Final Velocity will be \(v=0\)
Acceleration due to gravity \(g=-9.8\) (acting against the motion)
Height \(h\) =122.5 m
Let \(t\) be then time taken to reach maximum height
Time \(T\) to return on ground =\(2\times t\) \[ \begin{aligned} h&=\frac{1}{2}gt^2\\\\ 122.5&=\frac{1}{2}gt^2\\\\ 122.5&=\frac{1}{2}9.8t^2\\\\ 122.5&=4.9t^2\\\\ \implies t^2&=\frac{122.5}{4.9}\\\\ t^2&=25\\ t&=\sqrt{25}\\ &=5\,s\\ \implies T&=2t\\ T&=2\times 5\\ T&=10\,s \end{aligned} \]

Q14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Solution:
Height of the tower from which the stone is released \(=19.6\)
Initial velocity \(u=0\)
Final Velocity \(v=\) \[ \begin{aligned} v^2&=u^2+2gh\\ v^2&=0 + 2\times 9.8\times 19.6\\ v^2&=19.6\times 19.6\\ v&=\sqrt{19.6\times 19.6}\\&=19.6\,m/s \end{aligned} \]

Q15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Solution:
Initial velocity \(u=40\,m/s\)
Final Velocity \(v=0\)
Acceleration due to gravity \(g=1-0\,m/s^2\) against motion
Max height \(h=\) \[ \scriptsize \begin{aligned} v^2&=u^2+2ah\\ 0=&40^2+ 2\times (-10)\times h\\ \implies -2\times 10\times h&=-40\times40\\\\ h&=\frac{40\times 40}{2\times 10}\\\\ h&=80\,m \end{aligned} \] Total distance covered by stone on reaching the ground =\[2\times h\\=2\times 80\\=160\,m\] Displacement =0

Q16. Calculate the force of gravitation between the Earth and the Sun, given that the mass of the earth = \(6 \times 10^{24}\) kg and of the Sun = \(2 \times 10^{30}\) kg. The average distance between the two is \(1.5 \times 10^{11}\) m.

Solution:
Mass of the Earth \((m_e)= 6 \times 10^{24}\) kg
Mass of the Sun \((m_s)= 2 \times 10^{30}\) kg
Distance between them is \((R)=1.5 \times 10^{11}\) m
\(G=6.67\times 10^{-11}\)

Force of gravitation \(F\) between the earth and the Sun= \[ F=G\frac{m_em_s}{R^2} \] Substituting the values \[ \scriptsize \begin{aligned} F&=G\frac{m_em_s}{R^2}\\\\ &=6.67\times 10^{-11}\frac{6 \times 10^{24}\times 2 \times 10^{30}}{(1.5 \times 10^{11})^2}\\\\ &=\frac{6.67\times6\times2}{1.5\times1.5}\times \frac{10^{-11}\times 10^{24}\times10^{30}}{10^{22}}\\\\ &=\frac{6.67\times6\times2}{1.5\times1.5}\times 10^{(-11+24+30-22)}\\\\ &=\frac{6.67\times6\times2}{1.5\times1.5}\times 10^{21}\\\\ &=\frac{80.04\times 10^{21}}{2.25}\\ &=3.57\times 10^{22} \end{aligned} \]

Q17. A stone is allowed to fall from the top of a tower 100 m high, and at the same time, another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

Solution:
Let they met at the height of \(x\,m\)
and time taken to reach at meeting point \(=t\,s\)
Distance traveled by downward stone \(=x\)
Then, Distance travelled by other stone travelling upward \(= 100 -x\,m\)
$$x=\frac{1}{2}gt^{2}\tag{1}$$ Time taken by the other stone $$100-x=25t-\frac{1}{2}gt^{2}\tag{2}$$ Substituting the value of s from Eqn (1) into Eqn (2) $$\begin{align}100-\dfrac{1}{2}gt^{2}&=25t-\frac{1}{2}gt^{2}\\\\ 100&=25t\\\\ \implies t&=\frac{100}{25}\\\\ &=4s\\\end{align}$$ Substituting \(t=4\,s\) in equation (1) $$\begin{aligned}x&=\frac{1}{2}gt^{2}\\\\ &=\dfrac{1}{2}\times9.8\times 4\times 4\\\\ &=78.4m\end{aligned}$$ Both stones will meet at a height of 78.4 meters from the top after 4 seconds

Q18. A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) Its position after 4 s.

Solution:
Time of flight = 6s
Time taken in one way flight = \(3\,s\)
Let the Maximum height (h) reached by the ball $$\begin{aligned}h&=\frac{1}{2}gt^{2}\\ h&=\dfrac{1}{2}\times 9.8\times 3\times 3\\ &=4.9\times 9\\ &=44.1\,m\end{aligned}$$ Maximum Height that ball will cover is \(44.1\,m\)

Let the velocity by which the ball is thrown be \(u\) $$\scriptsize\begin{aligned}v^{2}-u^{2}&=2gh\\ 0^{2}-u^{2}&=2\times (-9.8)\times 44.1\\ u^{2}&=2\times 9.8\times 44.1\\ u^{2}&=2\times 4.9\times 2\times 49\times 9\\ u&=\sqrt{2\times 2\times 4.9\times 4.9\times 3\times 3}\\ &=2\times 4.9\times 3\\ &=9.8\times 3\\ &=29.4\,m/s\end{aligned}$$ Velocity at which ball is thrown is \(29.4\,m/s\)

At 4 Seconds, the ball will be on return flight and would have completed its flight of is therefore, height covered in \(1\,s\) in downward direction $$\begin{aligned}h&=\dfrac{1}{2}gt^{2}\\ &=\dfrac{1}{2}\times 9.8\times 1\times 1\\ &=4.9\end{aligned}$$ Ball will reach \(4.9\,m\) from from the top at \(4\,s\) of its flight

Q19. In what direction does the buoyant force on an object immersed in a liquid act?

Solution:
The buoyant force on an object dipped in a liquid always pushes straight upward. This upward push acts against the pull of gravity, helping objects float or feel lighter in the liquid.

Q20. Why does a block of plastic released under water come up to the surface of the water?

Solution:
A block of plastic rises to the surface when released underwater because it is less dense than the water around it.

When placed underwater, the plastic experiences an upward push called buoyant force, which works against gravity.

Since this upward force is stronger than the plastic’s weight, it is pushed upward and floats to the surface.

Essentially, the plastic’s lower density compared to water makes it rise instead of sinking.

Q21. The volume of 50 g of a substance is \(20 cm^3\). If the density of water is \(1 g cm^{–3}\), will the substance float or sink?

Solution:
Mass of the substance \(=50gm\)
Volume of substance \(=20cm^3\)
Density of substance \(d\) \[ \begin{aligned} d=\frac{\text{mass}}{\text{volume}}\\\\ =\frac{50\,gm}{20\,cm^3}\\\\ =2.5\,gm\,cm^{-3} \end{aligned} \] Density of Water \(=1 \,g \,cm^{–3}\lt 2.5\,gm\,cm^{-3}\) that of substance.
therefore,
Substance will sink in water.

Q22. The volume of a \(500\, g\) sealed packet is \(350\, cm^3\). Will the packet floats or sinks in water if the density of water is \(1\, g\, cm^{–3}\)? What will be the mass of the water displaced by this packet?

Solution:
Mass of the packet\(=500\,gm\)
Volume of packet\(=350\,cm^3\)
Density of substance \(d\) \[ \begin{aligned} d&=\frac{\text{mass}}{\text{volume}}\\\\ &=\frac{500\,gm}{350\,cm^3}\\\\ &\approx1.3\,gm\,cm^{-3} \end{aligned} \] Density of Water \(=1 \,g \,cm^{–3}\le 1.3\,gm\,cm^{-3}\) that of packet.
therefore,
The packet will sink in water.

The mass of water displaced is 350 g.