GRAVITATION-Notes

Whenever two bodies exist anywhere in the universe, an invisible force tries to pull them together. This force is called gravitation. It doesn't matter whether the bodies are planets, stones, or even grains of dust; each attracts the other silently and gently. Though for small objects, this pull is so weak we hardly notice it, gravitation is the mighty force behind falling rain, rolling rocks, and celestial dances of planets and stars.

• Gravitation acts between all objects—no exceptions, from the largest star to the tiniest particle.
• What we commonly call "gravity" is just gravitation special to Earth: it's why everything falls down and why we feel weight.
• Gravitation keeps planets in orbit and is the reason for ocean tides and the movement of satellites.
• Isaac Newton explained this universal pull through his law of gravitation, stating every object attracts every other object with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

This law states that every two bodies, whether giant planets or tiny grains of dust, attract each other with a force. The strength of this force depends on two things: the mass of each object, and how far they are from each other. Specifically, the force is stronger if the objects are heavier and weaker if they are farther apart.

Mathematical Expression

Consider two objects with masses\(M\) and \(m\), separated by a distance \(d\). Let the force of attraction between two objects be \(F\). According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses. That is, \[ F\propto M\times m \tag{1} \] And the force between two objects is inversely proportional to the square of the distance between them, that is, \[ F\propto \dfrac{1}{d^2} \tag{2} \] Combining Equations (1) and (2), we get \[ F\propto \dfrac{M\times m }{d^2} \] or, \[ \boxed{F=G \dfrac{M\times m }{d^2}} \] where \(G\) is the constant of proportionality and is called the universal gravitation constant.

The SI unit of G is \(\mathrm{N\,m^2\,kg^{-2}}\)

Value of G is \[\boxed{\mathrm{ 6.673\times 10^{-11}N\,m^2\,kg^{-2}}}\]

The mass of the earth is \(6 \times 10^{24}\) kg and that of the moon is \(7.4 \times 10^{22}\) kg. If the distance between the earth and the moon is \(3.84\times 10^5\) km, calculate the force exerted by the Earth on the moon. (Take G = \(6.7 \times 10^{-11} N m^2 kg^{-2}\))

Solution:
Mass of the earth (say it \(M\)) =\(6 \times 10^{24}\) kg and
Mass of Moon (\(m\)) = \(7.4 \times 10^{22}\) kg
G = \(6.7 \times 10^{-11} N m^2 kg^{-2}\))
Distance from Earth to Moon (\(d\)) =\[\begin{aligned}&3.84\times 10^5 \mathrm{km}\\ =&3.84\times 10^5\times 10^3\mathrm{m}\\ =&3.8\times 10^8\mathrm{m}\end{aligned}\] Force \((F)\) exerted by the earth on the moon= \[ F=G \dfrac{M\times m }{d^2}\tag{1} \] Substituting Values of \(M,\,m,\,d \text{ and } G\) in Formula \((1)\) \[\tiny \begin{aligned} F&=6.7 \times 10^{-11}\left(\frac{6 \times 10^{24}\times 7.4 \times 10^{22}}{(3.8\times 10^8)^2}\right)\\\\ &=\frac{6.7\times 6\times 7.4\times 10^{-11}\times 10^{24} \times 10^{22}}{3.8\times 3.8 \times 10^8\times 10^8}\\\\ &=\frac{6.7\times 6\times 7.4\times 10^{(-11+24+22)}}{3.8\times 3.8\times 10^{(8+8)}}\\\\ &=\frac{6.7\times 6\times 7.4\times 10^{35}}{3.8\times 3.8\times 10^{16}}\\\\ &=\frac{6.7\times 6\times 7.4}{3.8\times 3.8}\times 10^{(35-16)}\\\\ &=\frac{6.7\times 6\times 7.4}{3.8\times 3.8}\times 10^{19}\\\\ &=\frac{297.48}{14.44}\times 10^{19}\\\\ &=20.60\times 10^{19}\\\\ &=2.06\times 10^{20}\,\mathrm{N}\\\\ \end{aligned} \]

The importance of the Universal Law of Gravitation can be highlighted as follows:

• It explains why all objects are attracted toward the Earth, enabling us to stay grounded.
• It describes the revolution of the moon around the Earth, due to Earth's gravitational pull.
• It accounts for the planets revolving around the Sun, governing the structure of the solar system.
• It clarifies the phenomenon of ocean tides caused by the gravitational forces of the moon and the Sun.
• It helps us understand how artificial satellites orbit the Earth and assists in their launching.
• It is fundamental to modern technologies like GPS and space exploration.
• It provides a unified explanation for diverse phenomena both on Earth and in the cosmos.
• This law forms the base for understanding gravitational interactions among celestial bodies, planets, stars, and galaxies.

Free fall is defined as the motion of a body falling towards the Earth under the influence of Earth's gravitational force only, without any other forces like air resistance acting on it. In free fall, the body accelerates uniformly at a constant rate called the acceleration due to gravity, denoted by \(g\), which is approximately \(9.8m/s^2\) near the Earth's surface. This means the speed of the body increases steadily as it falls. Free fall is a fundamental concept in physics that explains how objects move solely under gravity's pull.

Derivation of Acceleration due to Gravity \(g\)

From the second law of motion:
\[F=ma\] where \(a\) is acceleration and \(m\) is mass of a object
In case of a freely falling object of \(m\), acceleration will be equal to acceleration due to gravity \(\g),so \[F=mg\tag{1}\] and \[F=G\dfrac{Mm}{d^2}\tag{2}\] Equating Equations (1) and (2) \[ mg=G\dfrac{Mm}{d^2}\\\\ g=G\dfrac{M}{d^2} \] where M is the mass of the Earth, and d is the distance between the object and the Earth.

Let an object be on or near the surface of the earth. The distance d will be equal to R, the radius of the Earth. Thus, for objects on or near the surface of the earth, \[g=G\dfrac{M}{R^2}\]

\[g=G\dfrac{M}{R^2}\] Substituting values of \(G,\,M,\,R\)
\(G =6.67\times 10^{-11}\,N m^2 kg^{-2}\)
\(M=6\times 10^{24}\,kg\)
\(R=6.4\times 10^6\,m\) \[g=G\dfrac{M}{R^2} \] \[\scriptsize \begin{aligned} &=6.67\times 10^{-11}N m^2 kg^{-2}\times \dfrac{6\times10^{24}kg}{(6.4\times10^6m)^2}\\\\ &=\dfrac{6.67\times6 \times 10^{(-11+24)}}{6.4\times 6.4\times10^{(6+6)}}\\\\ &=\dfrac{6.67\times6 \times 10^{(13)}}{6.4\times 6.4\times10^{(12)}}\\\\ &=\dfrac{6.67\times6 \times 10^{(13-12)}}{6.4\times 6.4}\\\\ &=\dfrac{6.67\times6 \times 10}{6.4\times 6.4}\\\\ &=\dfrac{40.02 \times 10}{40.96}\\\\ &=9.77\\ &\approx9.8m/s^2 \end{aligned}\] Thus, the value of acceleration due to gravity of the earth, \(g = 9.8\,m s^{–2}\)

All object experiences acceleration during free fall. This acceleration experienced by an object is independent of its mass.

This means that all objects, hollow or solid, big or small, should fall at the same rate.

As \(g\) is constant near the earth, all the equations for the uniformly accelerated motion of objects become valid with acceleration \(a\) replaced by \(g\).

The equations are:

• \[v=u+gt\]
• \[s=ut+\frac{1}{2}\, gt^2\]
• \[v^2=u^2 + 2gs\]

A car falls off a ledge and drops to the ground in \(0.5 s\). Let \(g = 10\, m s^{–2}\) (for simplifying the calculations).
(i) What is its speed on striking the ground?
(ii) What is its average speed during the \(0.5\,s\)?
(iii) How high is the ledge from the ground?

Solution:
Given:
\(g=10\,m/s^2\)
Time to reach the ground \(=0.5\,s\)
Initial Velocity \(u=0\)
(i) Speed on striking the ground
\[ \begin{aligned} v&=u+gt\\ &=10\times 0.5\\ &=5\,m/s \end{aligned} \] (ii) Average speed during the \(0.5\,s\)
\[ \begin{aligned} v_{av}&=\frac{v+u}{2}\\ &=\frac{5+0}{2}\\ &=2.5\,m/s \end{aligned} \] (iii) Height\(h\) the ledge from the ground?
\[ \begin{aligned} v^2&=u^2+gh\\ 5^2&=0^2+2\times 10\times h\\ \implies h&=\dfrac{25}{20}\\ &=1.25 \end{aligned} \]

An object is thrown vertically upwards and rises to a height of \(10\, m\).
Calculate
(i) the velocity with which the object was thrown upwards and
(ii) the time taken by the object to reach the highest point.

Solution:
Given:
Height \(=10\,m\)
\(g=-9.8\,m/s^2\)
Final Velocity \(v=0\)
(i) the velocity with which the object was thrown upwards \[ \scriptsize \begin{aligned} v^2&=u^2+2gh\\ 0&=u^2+2\times(-9.8\,ms^{-2})\times10\,s\\ \implies u&=\sqrt{196}\\ u&=14\,m/s \end{aligned} \] (ii) the time taken by the object to reach the highest point. \[ \scriptsize \begin{aligned} v&=u+gt\\ 0&=14\,ms^{-1}+(-9.8\,ms^{-2})\times t\\ -14\,ms^{-1}&=(-9.8\,ms^{-2})\times t\\ \implies t&=\dfrac{14}{9.8}\\ &\approx 1.43\,s \end{aligned}\]

Mass is the amount of matter present in a body. It tells us how much "stuff" or substance an object contains.
The mass of an object remains the same no matter where it is, whether on Earth, the Moon, or in space.
It is measured in kilograms (kg) and is a basic property of matter that does not change with location.

Weight is the force with which the Earth pulls an object towards its center.
It depends on both the mass of the object and the acceleration due to gravity.
Weight changes if the object is taken to places where gravity is different, but mass remains the same.
The SI unit of weight is Newton (N), and it is measured using a spring balance.
Mathematically, weight is given by the formula:

Weight=Mass×Acceleration due to gravity \[\boxed{\mathrm{W=mg}}\]

Let the mass of an object be \(m\).
Let its weight on the moon be \(W_m\).
Let the mass of the moon be \(M_m\) and its radius be \(R_m\).

By applying the universal law of gravitation, the weight of the object on the moon will be \[ W_m=G\dfrac{M_m\times m}{R_m^2}\tag{1} \] Let the weight of the same object on the Earth be \(W_e\).
The mass of the earth is\(M_e\) and its radius is \(R_e\) \[ W_e=G\dfrac{M_e\times m}{R_e^2}\tag{2} \] \[\tiny \begin{array}{|l|l|l|} \hline\text{Celestial Body}&\text{Mass}(kg)&\text{Radius}(m)\\\hline \hline\text{Earth}&5.98\times10^{24}&6.37\times10^6\\\hline \hline\text{Moon}&7.36\times 10^{22}&1.74\times 10^6\\\hline \end{array} \] \[ \begin{align} \tiny W_m&\tiny = G\frac{M_m m}{R_m^2}\\\\ &\tiny =\tiny G\frac{7.36\times10^{22}\times m}{(1.74\times10^6)^2}\\\\ &\tiny =\tiny Gm\left(\frac{7.36}{1.74\times 1.74}\right)\times10^{(22-12)}\\\\ &\tiny =\tiny Gm\left(\frac{7.36}{1.74\times 1.74}\right)\times10^{10}\tag{3} \end{align} \] \[ \begin{align} \tiny W_e&\tiny =G\frac{M_e m}{R_e^2}\\\\ &\tiny =\tiny G\frac{5.98\times10^{24}\times m}{(6.37\times10^6)^2}\\\\ &\tiny =\tiny Gm\left(\frac{5.98}{6.37\times 6.37}\right)\times10^{(24-12)}\\\\ &\tiny=\tiny Gm\left(\frac{5.98}{6.37\times 6.37}\right)\times10^{10^2}\tag{4} \end{align} \] Dividing Equation (3) by (4)
\[ \tiny \begin{aligned} \frac{W_m}{W_e} &= \frac{Gm\left(\frac{7.36}{1.74 \times 1.74}\right) \times 10^{10}}{Gm\left(\frac{5.98}{6.37 \times 6.37}\right) \times 10^{12}} \\\\ &= \frac{7.36 \times 6.37 \times 6.37}{5.98 \times 1.74 \times 1.74 \times 100} \\\\ &= \frac{298.64}{18.105 \times 100} \\\\ &= 0.165 \\ &\approx \frac{1}{6} \end{aligned} \] \[\boxed{\frac{W_m}{W_e} = \frac{1}{6}}\]

Mass of an object is 10 kg. What is its weight on the Earth?

Solution:
Mass \((m)\)=10kg
Acceleration due to gravity \((g)=9.8m/s^2\)
\[\begin{aligned}W&=mg\\&=10\times 9.8\\&=98\,N\end{aligned}\]

An object weighs 10 N when measured on the surface of the Earth. What would be its weight when measured on the surface of the moon?

Solution:
We know that: \[\boxed{\frac{W_m}{W_e} = \frac{1}{6}}\] Substituting value \(W_e=10N\) \[\begin{aligned}\frac{W_m}{W_e} &= \frac{1}{6}\\ \frac{W_m}{10}&=\frac{1}{6}\\ W_m&=10\times\frac{1}{6}\\\\ &=1.67N\end{aligned}\]

Thrust is a force that acts perpendicular (at a right angle) to the surface of an object. It is the push or pull force applied directly on a surface.

For example, when you press a nail into a board, the force your hand applies on the nail in the direction perpendicular to the board is called thrust. The SI unit of thrust is Newton (N), which is the same as the unit of force.

Thrust plays an important role in everyday activities and in machines like rockets and aeroplanes, where it helps to move objects by pushing against air or other surfaces.

Pressure is the amount of force applied on a surface divided by the area of that surface. It tells us how much force acts on each unit of area. The more force you apply on a smaller area, the greater the pressure. Pressure is measured in pascal (Pa), where 1 pascal equals 1 newton of force per square meter. The formula for pressure is: \[\text{Pressure}=\frac{\text{Force}}{\text{Area}}\\\\\boxed{P=\frac{F}{A}}\] Pressure helps us understand why sharp objects can cut or pierce easily and why we sink in soft sand when standing, but not when lying down.

\[\text{Pressure}=\frac{\text{Thrust}}{\text{Area}}\]

Pressure in fluids is defined as the force exerted by the fluid per unit area on the surface in contact with it. This force acts equally in all directions at a point within the fluid. The pressure in a fluid increases with depth due to the weight of the fluid above. It depends on the density of the fluid, the depth, and the acceleration due to gravity.

Buoyancy is the upward force exerted by a fluid (liquid or gas) on an object that is either partially or fully immersed in it.

This force is also called upthrust. Due to this force, objects immersed in a fluid appear lighter than they actually are.

Buoyancy depends on the volume of the fluid displaced by the object and the density of the fluid.

It is the reason why objects either float or sink in a fluid. For example, a ship floats on water because the buoyant force acting on it is equal to or greater than its weight, whereas a solid iron nail sinks because its weight is more than the buoyant force acting on it.

Archimedes’ Principle states that:

When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it.

• The law of gravitation states that the force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. The law applies to objects anywhere in the universe. Such a law is said to be universal.
• Gravitation is a weak force unless large masses are involved.
• The force of gravity decreases with altitude. It also varies on the surface of the earth, decreasing from the poles to the equator.
• The weight of a body is the force with which the earth attracts it.
• The weight is equal to the product of mass and acceleration due to gravity.
• The weight may vary from place to place, but the mass stays constant.
• All objects experience a force of buoyancy when they are immersed in a fluid.
• Objects having a density less than that of the liquid in which they are immersed, float on the surface of the liquid. If the density of the object is more than the density of the liquid in which it is immersed, then it sinks in the liquid.