We all know that newton discovered gravitation when an apple from an apple tree fell in front of him. But why does it exist, and do we also exert gravitational force on each other? What is a gravitational field? How do the planets revolve around the sun? Do the planets exert gravitational force on each other? How does the speed of the planet vary in its orbit? What are Kepler’s laws?

Let us read further to discuss gravitation in detail and find the answers to these questions.

Newton’s Law of Gravitation

Gravitation is the phenomenon by which any two bodies that possess some mass to exert force on each other. This force exists due to the mass. Gravitational forces do not require physical contact to act. The gravitational force is always attractive and independent of the medium.

Newton discovered this force and gave the expression for the magnitude of the force; that is, the magnitude of the force is directly proportional to the product of the masses of the bodies and inversely proportional to the square of the distance between them.
\(F \propto {m_1} \times {m_2}\)
\(F \propto \frac{1}{{{r^2}}}\)
\( \Rightarrow F = \frac{{G{m_1}{m_2}}}{{{r^2}}}\)
Where,
\(G\) is the proportionality constant known as the universal gravitational constant.
\(\overrightarrow {{F_{1,2}}} = – \overrightarrow {{F_{2,1}}} \)
\(\overrightarrow {\left| {{F_{1,2}}} \right|} = \left| {\overrightarrow {{F_{2,1}}} } \right|\)
The limitation of Newton’s Law of gravitation is that it is only applicable for point sized objects.

Gravitational Field

We do not need any physical contact to apply force in the case of gravitational force. Any particle kept in space will experience a gravitation force due to objects around it, so we explain this phenomenon using the force field that is an object kept in space will generate a field around it. If any other object is kept in that field, it will experience force. Its unit is Newton per \(\rm{kg}\).

The intensity of the field is given by,
\(\overrightarrow E = \frac{{\overrightarrow F }}{{{m_0}}}\).

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Calculation of Field

Due to point mass
\(\left| {\overrightarrow E } \right| = \frac{{Gm}}{{{r^2}}}\)
Due to discrete distribution of mass
The total gravitational field at a point is given by the vector sum of the fields due to individual masses.

\(\overrightarrow E = \overrightarrow {{E_1}} + \overrightarrow {{E_2}} + \overrightarrow {{E_3}} + ….\)
Due to continuous distribution
The net gravitational field is given by the integration of the fields due to differential elements.

\(\overrightarrow E = \int {\overrightarrow {{\text{d}}E} } \)
Due to a thread

The component of the gravitational field in the direction perpendicular to the thread is,
\({E_ \bot } = \frac{{G\lambda }}{d}\left| {{\text{sin}}\left( {{{{\theta }}_1}} \right) + {\text{sin}}\left( {{{{\theta }}_2}} \right)} \right|\)
The component of the gravitational field in the direction parallel to the thread is,
\({E_\parallel } = \frac{{G\lambda }}{d}\left| {\cos \left( {{\theta _2}} \right) – \cos \left( {{\theta _1}} \right)} \right|\)
Where,
\(\theta _1\) and \(\theta _2\) are the angles subtended by the ends of the thread at that point
\(d\) is the perpendicular distance of that point from the thread.
\(\lambda\) is the mass per unit length of the thread.
If \(\theta _1 = \theta _2\), then the component of the gravitational field in the direction parallel to the thread will be zero.
Due to infinite uniform thread

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\(\overrightarrow E = \frac{{2G\lambda }}{{\text{d}}}\)
In the case of an infinite thread, the component of the gravitational field in the direction parallel to the thread will be zero.
The gravitational field in the direction parallel to the length of the thread is cancelled out due to symmetry, and only the component of the Gravitational field in the direction perpendicular to the thread exists.
Due to semi-infinite thread

\({E_ \bot } = \frac{{G\lambda }}{{\text{d}}}\)
\({E_\parallel } = \frac{{G\lambda }}{{\text{d}}}\)
We can get the above result by putting in the value of \({\theta _1} = \frac{\pi }{2}\) and \({\theta _2} = 0\)
Due to uniform circular arc at its centre

\(\overrightarrow E = \frac{{2G\lambda }}{R}\sin \left( {\frac{\theta }{2}} \right)\)
Where,
\(\lambda\) is the mass per unit length of the thread.
\(R\) is the radius of the arc.
\(\theta\) is the angle subtended by the arc at its centre.
Due to uniform circular ring at its axis

\(\overrightarrow E = \frac{{GMx}}{{{{\left( {{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\)
Where,
\(m\) is the mass of the ring.
\(x\) is the distance of the point on its axis from the centre of the ring.
Due to uniform disc at a point on its axis

\(\overrightarrow E = G\sigma 2\pi \left( {1 – \cos \left( \theta \right)} \right)\)
Where,
\(\theta\) is the angle subtended by the radius of the disc on that point.
\(\sigma\) is the mass per unit area of the disc.
Due to infinite sheet

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\(G\sigma 2\pi \)
The gravitational field due to a uniform hollow sphere or shell.

1. Outside or on the boundary of the shell
\(r \geqslant R\)
\( \Rightarrow E = \frac{{GM}}{{{r^2}}}\)
2. Inside the shell
\(r < R\)
The gravitational field inside the shell will be zero.

The gravitational field due to solid sphere,
1. Outside of the boundary of the sphere
\(r \geqslant R\)
\( \Rightarrow E = \frac{{GM}}{{{r^2}}}\)
2. Inside the sphere
\(r < R\)
Now, since the sphere is uniform, the density is given to be,
\(\rho = \frac{M}{{4\pi {R^2}}}\)
Therefore, the mass enclosed by the sphere of radius \(r\) will be,
\({M_{{\text{enclosed}}}} = \frac{{3M}}{{4\pi {R^3}}} \times \frac{4}{3}\pi {r^3}\)
\( \Rightarrow E = \frac{{GMr}}{{{R^3}}}\)

Gravitational Potential

The gravitational potential difference between two points, ‘\(A\)’ and ‘\(B\)’, is defined as the work done per unit mass in moving a mass from point ‘\(A\)’ to ‘\(B\)’ slowly. It SI unit is Joules per kilogram.
\({V_B} – {V_A} = \frac{{{W_{\rm{ext}}}}}{{{m_0}}}\left( {A \to B,\,{\text{slowly}}} \right)\)
\({W_{\rm{ext}}}\left( {A \to B} \right) = \int_A^B {\overrightarrow F } \cdot \overrightarrow {dr} \)
\({W_{\rm{ext}}}\left( {A \to B} \right) = \int_A^B { – {m_0}\overrightarrow E } \cdot \overrightarrow {dr} \)
\({V_B} – {V_A} = \int_A^B {\overrightarrow E } \cdot \overrightarrow {dr} \)

Here the point \(A\) is the reference point or initial point.
Conventionally we consider the point at infinity to be the reference point.
Thus, we can say that the gravitational potential of a point is defined as work done per unit mass to bring a mass from infinity to that point.
Also, generally, the potential at infinity is considered to be zero,
\({V_\infty } = 0\)
Therefore, the expression for gravitational potential comes out to be,

\({V_p} = {V_p} – {V_\infty } = – \int_\infty ^p {\overrightarrow E } \cdot \overrightarrow {dr} \)
From the above definition, we can also infer that the magnitude of the gravitational field in a given direction is equal to the rate of change of potential with respect to distance,
\(\left| {\overrightarrow E } \right| = \frac{{dV}}{{dr}}\).

Calculation of Gravitational Potential

Due to a point mass
A point mass is present in the space, and we need to calculate the potential at a point at a distance ‘\(r\)’, near the mass.
We know that the gravitational potential of a point is given by,
\({V_p} = {V_p} – {V_\infty } = – \int_\infty ^p {\overrightarrow E } \cdot \overrightarrow {dr} \)
Conventionally,
\({V_\infty } = 0\)
\({V_p} = – \int_\infty ^P {\overrightarrow E } \cdot \overrightarrow {dr} \)
Putting the value of gravitational field due to a point mass we get,
\({V_p} = – \int_\infty ^r {\frac{{Gm}}{{{r^2}}}} \cdot \overrightarrow {dr} \)
\({V_p} = – \frac{{Gm}}{r}\)
Due to the distribution of masses.
Potential is a scalar quantity, and therefore the potential due to a distribution of masses is given by the scalar sum of the potential due to individual masses.

\({V_P} = {V_1} + {V_2} + {V_3}\)
Due to continuous mass bodies
For continuous bodies, we get the potential by integrating the potential due to differential elements.

\({V_P} = \smallint dV\)
Gravitational potential due to a thread

\(V = – G\lambda .{\text{ln}}\left( {\frac{{{\text{sec}}\left( {{\theta _1}} \right) + {\text{tan}}\left( {{\theta _1}} \right)}}{{{\text{sec}}\left( {{\theta _2}} \right) – {\text{tan}}\left( {{\theta _2}} \right)}}} \right)\)
Where,
\(\theta _1\) and \(\theta _2\) are the angle subtended by the ends of the thread at that point, as shown in the figure.
\(\lambda\) is the mass per unit length of the thread.

For an infinite thread, the potential at infinity is not zero. In that case, the potential of a point is not defined, but the potential difference between two points can be calculated which is given by,
\({V_B} – {V_A} = – 2G\lambda .{\text{ln}}\left( {\frac{{{r_A}}}{{{r_B}}}} \right)\)
Where,
\(r_A\) and \(r_B\) are the distance of the two points from the thread.
Potential due to a circular arc

\({V_p} = – \frac{{GM}}{R}\)
\(R\) is the radius of the arc.
\(Q\) is the mass of the arc.
Potential due to a ring at a point on its axis.

\({V_p} = – \frac{{GM}}{{\sqrt {{R^2} + {x^2}} }}\)
Where,
\(M\) is the mass on the ring.
\(x\) is the distance of the point on its axis from the centre of the ring.
\(R\) is the radius of the ring
Potential due to uniform disc at a point on its axis

\(V = – 2\pi \sigma G\left( {\sqrt {{R^2} + {x^2}} – x} \right)\)
Where,
\(\sigma\) is the area mass density of the disc.
\(x\) is the distance of the point on its axis from the centre of the disc.
\(R\) is the radius of the disc
Gravitational potential due to uniform shell
1. Outside the shell
\(r > R\)
\({V_p} = – \frac{{GM}}{r}\)
2. On the boundary of the shell
\(r = R\)
\({V_p} = – \frac{{GM}}{r}\)
3. Inside the shell
\(r < R\)
\({V_p} = – \frac{{GM}}{R}\)
The net gravitational field inside the shell is zero; therefore, the potential remains constant inside the shell.

Gravitational potential due to a uniform sphere.
1. Outside the shell
\(r > R\)
\({V_p} = – \frac{{GM}}{r}\)
2. On the boundary of the shell
\(r = R\)
\({V_p} = – \frac{{GM}}{R}\)
3. Inside the shell
\(r < R\)
\({V_p} = – \frac{{GM}}{{{R^3}}}\left[ {\frac{3}{2}{R^2} – \frac{{{r^2}}}{2}} \right]\).

Gravitational Potential Energy

Gravitational potential energy can be defined as the work done by an external agent in changing the system’s configuration slowly.
Example: Three masses \(m_1\), \(m_2\) and \(m_3\) are placed in space, and we need to calculate the gravitational potential energy of the system.
At first, we bring the first mass from infinity to origin; since there is no gravitational field in space, the work done required to bring the first mass will be zero.
For the second mass, since the gravitational field is present due to the first mass, the work done to bring it from infinity to a point at a distance \(r_{12}\) is given by,
\({W_{\rm{ext}}} = mV\)
\( \Rightarrow {W_{\rm{ext}}} = {m_2}\frac{{ – G{m_1}}}{{{r_{12}}}} = – \frac{{G{m_1}{m_2}}}{{{r_{12}}}}\)
For the third mass, we have a gravitational field due to two masses \(m_1\) and \(m_2\) present in space, thus work done in bringing the mass from the infinity to that point will be,
\({W_{\rm{ext}}} = mV\)
\( \Rightarrow {W_{\rm{ext}}} = – \frac{{G{m_1}{m_3}}}{{{r_{13}}}} – \frac{{G{m_2}{m_3}}}{{{r_{23}}}}\)
Therefore, the total work done will be,
\(U = – \frac{{G{m_1}{m_2}}}{{{r_{12}}}} – \frac{{G{m_1}{m_3}}}{{{r_{13}}}} – \frac{{G{m_2}{m_3}}}{{{r_{23}}}}\)
Thus, we can infer that the gravitational potential energy for distribution of mass will be,
\(U = \mathop {\sum – \frac{{G{m_i}{m_j}}}{{{r_{ij}}}}}\limits_{i \ne j} \)
For continuous bodies, we also have self-energy apart from interaction energy.
Self-energy means the amount of work required to build the continuous body while bringing the elemental masses from infinity.
The self-energy of a spherical shell is,
\(U = – \frac{{G{M^2}}}{{2R}}\)
The self-energy of a solid sphere is,
\(U = – \frac{{3G{M^2}}}{{5R}}\).

Kepler’s Laws of Planetary Motion

All planets move in an elliptical path with the sun at one of its foci.
The areal velocity of the planets remains constant.

\(\frac{{dA}}{{dt}} = \frac{L}{{2m}} = {\text{constant}}\)
Where,
\(L\) is the angular momentum of the planet about the sun.
The square of the time period of the planet is directly proportional to the cube of the semi-major axis.

Energy of an Orbiting Satellite

The total energy of a satellite orbiting the Earth at \({\text{h}}\) distance from the Earth’s surface is calculated as follows:
The satellite’s total energy is the sum of its kinetic energy and gravitational potential energy. The satellite’s potential energy is,
\({\text{U}} = – \frac{{{\text{G}}{{\text{M}}_{\text{s}}}{{\text{M}}_{\text{E}}}}}{{\left( {{{\text{R}}_{\text{E}}} + {\text{h}}} \right)}}\)
Here \({{\text{M}}_{\text{s}}} = \) mass of the satellite, \({{\text{M}}_{\text{E}}} = \) mass of the Earth, \({{\text{R}}_{\text{E}}} = \) radius of the Earth.
The Kinetic energy of the satellite is given by the equation:
\({\text{K}}{\text{.E}} = \frac{1}{2}{{\text{M}}_{\text{s}}}{{\text{v}}^2}\)
Here \({\text{v}}\) is the orbital speed of the satellite and is equal to
\({\text{v}} = \sqrt {\frac{{{\text{G}}{{\text{M}}_{\text{E}}}}}{{\left( {{{\text{R}}_{\text{E}}} + {\text{h}}} \right)}}} \)
Substituting the value of \({\text{v}}\) in the above equation, the kinetic energy of the satellite becomes,
\({\text{K}}{\text{.E}} = \frac{1}{2}\frac{{{\text{G}}{{\text{M}}_{\text{E}}}{{\text{M}}_{\text{s}}}}}{{\left( {{{\text{R}}_{\text{E}}} + {\text{h}}} \right)}}\)
Therefore the total energy of the satellite is
\({\text{E}} = \frac{1}{2}\frac{{{\text{G}}{{\text{M}}_{\text{E}}}{{\text{M}}_{\text{s}}}}}{{\left( {{{\text{R}}_{\text{E}}} + {\text{h}}} \right)}} – \frac{{{\text{G}}{{\text{M}}_{\text{s}}}{{\text{M}}_{\text{E}}}}}{{\left( {{{\text{R}}_{\text{E}}} + {\text{h}}} \right)}}\)
\({\text{E}} = – \frac{{{\text{G}}{{\text{M}}_{\text{s}}}{{\text{M}}_{\text{E}}}}}{{2\left( {{{\text{R}}_{\text{E}}} + {\text{h}}} \right)}}\)
The negative sign in the total energy indicates that the satellite is bound to the Earth and cannot escape. The total energy tends to zero as \({\text{h}}\) approaches \(\infty \). It means that the satellite is completely of the influence of Earth’s gravity and is not physically bound to Earth over long distances.

Sample Problems

Q.1. A satellite is moving with velocity ‘\(v\)’ in orbit with radius ‘\(r\)’. Find the relation between the time period of the revolution and the radius of the orbit.
Ans: Given,
The velocity of the satellite is ‘\(v\)’.
The radius of the orbit is ‘\(r\)’.
The gravitational attraction provides centripetal acceleration.
\(\frac{{GMm}}{{{r^2}}} = m\frac{{{v^2}}}{r}\)
\( \Rightarrow {v^2} = \frac{{GM}}{r}\)
The time period is given by,
\(T = \frac{{2\pi r}}{v}\)
\( \Rightarrow T = \frac{{2\pi r}}{{\sqrt {\frac{{GM}}{r}} }}\)
\( \Rightarrow {T^2} = \frac{{4{\pi ^2}r3}}{{GM}}\).

Q.2. Find the escape velocity of a satellite moving with velocity ‘\(v\)’ in orbit with radius ‘\(r\)’.
Ans: For a bounded system like planet-satellite or like an atom, the total energy is less than zero, and for an unbounded system, the total energy will be zero or more than zero.
The total energy of the satellite is given by the sum of kinetic and potential energy.
The kinetic energy of the satellite is,
\(\frac{1}{2}\frac{{GMm}}{r}\)
The potential energy of the satellite is given by,
\( – \frac{{GMm}}{r}\)
For the case of escape velocity, the net energy will be zero that is,
\({\text{TE}} = \frac{1}{2}m{v_e}^2 – \frac{{GMm}}{r} = 0\)
\( \Rightarrow {v_e} = \sqrt {\frac{{2GM}}{r}} \).

Summary

In this article, we discussed gravitation and newton’s law of gravitation. We learnt that gravitational force is directly proportional to the product of masses and inversely proportional to the square of the radius. We also discussed the gravitational field and also learnt the expression of the field due to various objects. We also calculated the gravitational potential due to point mass and different continuous bodies. At last, we defined the three laws of planetary motion given by Kepler.

FAQs

Q.1. What is gravitation?
Ans: Gravitation is a phenomenon in which any two bodies attract each other due to their mass.

Q.2. Can gravitation force be repulsive?
Ans: Gravitational force can not be repulsive. It is only attractive.

Q.3. What is a gravitational field?
Ans: A gravitational field is a force field generated by any particle with some mass, and if some other mass is present in that field, it will experience some force.

Q.4. What is gravitational potential?
Ans: The gravitational potential of a point is defined as work done per unit mass in bringing a mass from infinity to that point.

Q.5. What is Kepler’s law of planetary motion?
Ans: Kepler gave three laws for planetary motion.
According to Kepler’s first law of planetary motion, all planets move in an elliptical orbit with the sun at one of its foci.
Kepler’s second law states that in orbit, the areal velocity of the planet remains constant.
Kepler’s third law states that the square of the time period is directly proportional to the cube of the semi-major axis.

Q.6. What is gravity?
Ans: Gravity is the constant acceleration experienced by anybody due to the earth’s gravitational field.

We hope you find this article on ‘Gravitation‘ helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them.