When a body of mass (m) is moved from infinity to a point inside the gravitational influence of a source mass (M) without accelerating it, the amount of work done in displacing it into the source field is stored in the form of potential energy. This is known as gravitational potential energy. It is represented by the symbol Ug.

Explanation: We know that the potential energy of a body at a given position is defined as the energy stored in the body at that position. If the position of the body changes due to the application of external forces the change in potential energy is equal to the amount of work done on the body by the forces.

Under the action of gravitational force, the work done is independent of the path taken for a change in position so the force is a conservative force. Besides, all such forces have some potential in them. The gravitational influence on a body at infinity is zero, therefore, potential energy is zero, which is called a reference point.

Gravitational Potential Energy Formula

The equation for gravitational potential energy is:  \(GPE = m \times g \times h\)

Where m is the mass in kilograms, g is the acceleration due to gravity (\(9.8 \mathrm{~m} \mathrm{~s}^{-2}\) on Earth) and h is the height above the ground in meters.

Derivation of Gravitational Potential Energy Equation

Consider a source mass ‘ \(M\) ‘ is placed at a point along the \(x\)-axis, initially, a test mass ‘ \(m\) ‘ is at infinity. A small amount of work done in bringing it without acceleration through a very small distance \((\mathrm{dx})\) is given by \(\mathrm{dw}=\mathrm{Fdx}\)

Here, \(\mathrm{F}\) is an attractive force and the displacement is towards the negative \(\mathrm{x}\)-axis direction so \(\mathrm{F}\) and \(\mathrm{dx}\) are in the same direction. Then,

\(
\mathrm{dW}=\left(\mathrm{GMm} / \mathrm{x}^2\right) \mathrm{dx}
\)
Integrating on both sides
\(
\begin{aligned}
&W=\int_{\infty}^r \frac{G M m}{x^2} d x \\
&W=-\left[\frac{G M m}{x}\right]_{\infty}^r \\
&W=-\left[\frac{G M m}{r}\right]-\left(\frac{G M m}{\infty}\right)
\end{aligned}
\)
\(
W=-\frac{G M m}{r}
\)

Since the work done is stored as its potential energy \(U\), therefore gravitational potential energy at a point which is at a distance ‘ \(r\) ‘ from the source mass is given by;
\(
{U(r)}=-\frac{G M m}{r}
\)

If a test mass moves from a point inside the gravitational field to the other point inside the same gravitational field of source mass, then the change in potential energy of the test mass is given by;
\(
\Delta U= U\left(r_2\right)-U\left(r_1\right)={GMm}\left(1 / \mathrm{r}_{\mathrm{1}}-1 / \mathrm{r}_{\mathrm{2}}\right)
\)
If \(\mathrm{r}_{\mathrm{1}}>\mathrm{r}_{\mathrm{2}}\) then \(\Delta {U}\) is negative.

The gravitational potential energy of a two-particle two-particle system

Let a particle of mass \(m_1\) be kept fixed at a point \(A\) (figure below) and another particle of mass \(m_2\) is taken from a point \(B\) to a point \(C\). Initially, the distance between the particles is \(A B=r_1\) and finally, it becomes \(A C=r_2\). We have to calculate the change in potential energy of the system of the two particles as the distance changes from \(r_1\) to \(r_2\).

Consider a small displacement when the distance between the particles changes from \(r\) to \(r+d r\). In the figure, this corresponds to the second particle going from \(D\) to \(E\).
The force on the second particle is
\(
F=\frac{G m_1 m_2}{r^2} \text { along } \overrightarrow{D A} \text {. }
\)
The work done by the gravitational force in the displacement is

\(
d W=-\frac{G m_1 m_2}{r^2} d r .
\)
The increase in potential energy of the two-particle system during this displacement is
\(
d U=-d W=\frac{G m_1 m_2}{r} d r .
\)
The increase in potential energy as the distance between the particles changes from \(r_1\) to \(r_2\) is
\(
\begin{aligned}
&U\left(r_2\right)-U\left(r_1\right)=\int d U \\
&=\int_{r_1}^{r_2} \frac{G m_1 m_2}{r^2} d r=G m_1 m_2 \int_{r_1}^{r_2} \frac{1}{r^2} d r \\
&=G m_1 m_2\left[-\frac{1}{r}\right]_{r_1}^{r_2} \\
&=G m_1 m_2\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \dots(1.1)
\end{aligned}
\)
We choose the potential energy of the two-particle system to be zero when the distance between them is infinity. This means that we choose \(U(\infty)=0\). By (1.1) the potential energy \(U(r)\), when the separation between the particles is \(r\), is
\(
\begin{aligned}
U(r) &=U(r)-U(\infty) \\
&=G m_1 m_2\left[\frac{1}{\infty}-\frac{1}{r}\right]=-\frac{G m_1 m_2}{r} .
\end{aligned}
\)
The gravitational potential energy of a two-particle system is
\(
U(r)=-\frac{G m_1 m_2}{r} \dots(1.2)
\)
where \(m_1\) and \(m_2\) are the masses of the particles, \(r\) is the separation between the particles and the potential energy is chosen to be zero when the separation is infinite.

Example 1: Find the work done in bringing three particles, each having a mass of \(100 \mathrm{~g}\), from large distances to the vertices of an equilateral triangle of side \(20 \mathrm{~cm}\).

Solution: When the separations are large, the gravitational potential energy is zero. When the particles are brought at the vertices of the triangle \(A B C\), three pairs \(A B, B C\) and \(C A\) are formed. The potential energy of each pair is \(-G m_1 m_2 / r\) and hence the total potential energy becomes
\(
\begin{aligned}
U &=3 \times\left[-\frac{G m_1 m_2}{r}\right] \\
&=3 \times\left[-\frac{6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2 \times(0.1 \mathrm{~kg}) \times(0.1 \mathrm{~kg})}{0.20 \mathrm{~m}}\right] \\
&=-1.0 \times 10^{-11} \mathrm{~J} .
\end{aligned}
\)
The work done by the gravitational forces is \(W=-U=1.0 \times 10^{-11} \mathrm{~J}\). If the particles are brought by some external agency without changing the kinetic energy, the work done by the external agency is equal to the change in potential energy \(=-1.0 \times 10^{-11} \mathrm{~J}\).

Example 2: Find the potential energy of a system of four particles placed at the vertices of a square of side \(l\). Also, obtain the potential at the centre of the square.

Solution:

Consider four masses each of mass \(\mathrm{m}\) at the corners of a square of side \(l\); See FIgure above. We have four mass pairs at distance \(l\) and two diagonal pairs at distance \(\sqrt{2} l\)
Hence,
\(
W(r)=-4 \frac{G m^2}{l}-2 \frac{G m^2}{\sqrt{2} l}
\)
\(
=-\frac{2 G m^2}{l}\left(2+\frac{1}{\sqrt{2}}\right)=-5.41 \frac{G m^2}{l}
\)
The gravitational potential at the centre of the square \((r=\sqrt{2} l / 2)\) is
\(
U(r)=-4 \sqrt{2} \frac{\mathrm{G} \mathrm{m}}{l} \text {. }
\)

Expression for Gravitational Potential Energy at Height \((\mathrm{h})\) – Derive \(\Delta {U}=\mathrm{mgh}\)

If a body is taken from the surface of the earth to a point at a height ‘ \(h\) ‘ above the surface of the earth, then \(r_1=r\) and \(r_2=r+h\) then,
\(
\begin{aligned}
&\Delta {U}={GMm}[1 / {r}-1 /({r}+{h})] \\
&\Delta {U}={GMmh} / {r}({r}+{h})
\end{aligned}
\)
When, \(h \ll r\), then, \({r}+{h}={r}\) and \({g}={GM} / {r}^2\).
On substituting this in the above equation we get,
Gravitational Potential Energy \(\Delta {U}=mgh\)

Note:

• The weight of a body at the centre of the earth is zero due to the fact that the value of g at the centre of the earth is zero.
• At a point in the gravitational field where the gravitational potential energy is zero, the gravitational field is zero.

What is Gravitational Potential?

The amount of work done in moving a unit test mass from infinity into the gravitational influence of source mass is known as gravitational potential.
Simply, it is the gravitational potential energy possessed by a unit test mass
\(
\begin{aligned}
&{V}={U} / {m} \\
&{V}=-{GM} / {r}
\end{aligned}
\)
Important Points to remember:

• The gravitational potential at a point is always negative, \({V}\) is maximum at infinity.
• The SI unit of gravitational potential is \({J} / {Kg}\).
• The dimensional formula is \(M^0 L^2 T^{-2}\).

Calculation Of Gravitational Potential

(A) Derive Potential due to a Point Mass

Suppose a particle of mass \(M\) is kept at a point \(A\) (figure below) and we have to calculate the potential at a point \(P\) at a distance \(r\) away from \(A\). The reference point is at infinity.

The potential at the point \(P\) is
\(
V(r)=\frac{U(r)-U(\infty)}{m} .
\)
But \(U(r)-U(\infty)=-\frac{G M m}{r}\)
so that,
\(
V=-\frac{G M}{r} 
\)
The gravitational potential due to a point mass \(M\) at a distance \(r\) is \(-\frac{G M}{r}\).

(B) Potential due to a Uniform Ring at A Point on its Axis

Let the mass of the ring be \(M\) and its radius be \(a\). We have to calculate the gravitational potential at a point \(P\) on the axis of the ring (figure below). The centre is at \(O\) and \(O P=r\).

Consider any small part of the ring of mass \(d m\). The point \(P\) is at a distance \(z=\sqrt{a^2+r^2}\) from \(d m\).
The potential at \(P\) due to \(d m\) is
\(
d V=-\frac{G d m}{z}=-\frac{G d m}{\sqrt{a^2+r^2}} .
\)
The potential \(V\) due to the whole ring is obtained by summing the contributions from all the parts. As the potential is a scalar quantity, we have

\(
\begin{aligned}
V &=\int d V \\
&=\int-\frac{G d m}{\sqrt{a^2+r^2}} \\
&=-\frac{G}{\sqrt{a^2+r^2}} \int d m \\
&=-\frac{G M}{\sqrt{a^2+r^2}} .
\end{aligned}
\)
In terms of the distance \(z\) between the point \(P\) and any point of the ring, the expression for the potential is given by
\(
V=-\frac{G M}{z}
\)

(C) Potential due to a Uniform Thin Spherical Shell

Let the mass of the given spherical shell be \(M\) and the radius \(a\). We have to calculate the potential due to this shell at a point \(P\). The centre of the shell is at \(O\) and \(O P=r\) (figure below). 

Let us draw a radius \(O A\) making an angle \(\theta\) with \(O P\). Let us rotate this radius about \(O P\) keeping the angle \(A O P\) fixed at value \(\theta\). The point \(A\) traces a circle on the surface of the shell. Let us now consider another radius at an angle \(\theta+d \theta\) and likewise rotate it about \(O P\). Another circle is traced on the surface of the shell. The part of the shell included between these two circles (shown shaded in the above figure) may be treated as a ring.

The radius of this ring is \(a \sin \theta\) and hence the perimeter is \(2 \pi a \sin \theta\). The width of the ring is \(a d \theta\). The area of the ring is
\(
\begin{aligned}
&(2 \pi a \sin \theta)(a d \theta) \\
=& 2 \pi a^2 \sin \theta d \theta .
\end{aligned}
\)
The total area of the shell is \(4 \pi a^2\). As the shell is uniform, the mass of the ring enclosed is
\(
\begin{aligned}
d m &=\frac{M}{4 \pi a^2}\left(2 \pi a^2 \sin \theta d \theta\right) \\
&=\frac{M}{2} \sin \theta d \theta .
\end{aligned}
\)
Let the distance of any point of the ring from \(P\) be \(A P=z\). From the triangle \(O A P\)
\(\begin{aligned} z^2 &=a^2+r^2-2 a r \cos \theta \\ \text { or, } & 2 z d z =2 a r \sin \theta d \theta \\ \text { or, } & \sin \theta d \theta =\frac{z d z}{a r} . \end{aligned}\)
Thus, the mass of the ring is
\(
d m=\frac{M}{2} \sin \theta d \theta=\frac{M}{2 a r} z d z .
\)
As the distance of any point of the ring from \(P\) is \(z\), the potential at \(P\) due to the ring is
\(
\begin{aligned}
d V &=-\frac{G d m}{z} \\
&=-\frac{G M}{2 a r} d z .
\end{aligned}
\)
As we vary \(\theta\) from 0 to \(\pi\), the rings formed on the shell cover up the whole shell. The potential due to the whole shell is obtained by integrating \(d V\) within the limits \(\theta=0\) to \(\theta=\pi\).

Case 1: \(P\) is outside the shell \((r>a)\)

As figure shows, when \(\theta=0\), the distance \(z=A P=r-a\). When \(\theta=\pi\), it is \(z=r+a\). Thus, as \(\theta\) varies from 0 to \(\pi\), the distance \(z\) varies from \(r-a\) to \(r+a\). Thus,
\(
\begin{aligned}
V &=\int d V=-\frac{G M}{2 a r} \int_{r-a}^{r+a} d z \\
&=-\frac{G M}{2 a r}[z]_{r-a}^{r+a} \\
&=-\frac{G M}{2 a r}[(r+a)-(r-a)] \\
&=-\frac{G M}{r} \dots(1.3)
\end{aligned}
\)
To calculate the potential at an external point, a uniform spherical shell may be treated as a point particle of equal mass placed at its centre.

Case 2: \(P\) is inside the shell \((r<a)\)

In this case when \(\theta=0\), the distance \(z=A P\) \(=a-r\) and when \(\theta=\pi\) it is \(z=a+r\) (figure below). Thus, as \(\theta\) varies from 0 to \(\pi\), the distance \(z\) varies from \(a-r\) to \(a+r\).

Thus, the potential due to the shell is
\(
\begin{aligned}
V &=\int d V \\
&=-\frac{G M}{2 a r}[z]_{a-r}^{a+r} \\
&=-\frac{G M}{2 a r}[(a+r)-(a-r)] \\
&=-\frac{G M}{a} \dots(1.4)
\end{aligned}
\)

This does not depend on \(r\). Thus, the potential due to a uniform spherical shell is constant throughout the cavity of the shell.

The figure below shows graphically the variation of potential (drawn using equation 1.4) with the distance from the centre of the shell.

Example 3: A particle of mass \(M\) is placed at the centre of a uniform spherical shell of equal mass and radius a. Find the gravitational potential at a point \(P\) at a distance a \(/ 2\) from the centre.

Solution: The gravitational potential at the point \(P\) due to the particle at the centre is
\(
V_1=-\frac{G M}{a / 2}=-\frac{2 G M}{a} \text {. }
\)
The potential at \(P[latex] due to the shell is
[latex]
V_2=-\frac{G M}{a} \text {. }
\)
The net potential at \(P\) is \(V_1+V_2=-\frac{3 G M}{a}\).

(D) Potential due to a Uniform Solid Sphere

The situation is shown in the figure below. Let the mass of the sphere be \(M\) and its radius \(a\). We have to calculate the gravitational potential at a point \(P\). Let \(O P=r\).

Let us draw two spheres of radii \(x\) and \(x+d x\) concentric with the given sphere. These two spheres enclose a thin spherical shell of volume \(4 \pi x^2 d x\). The volume of the given sphere is \(\frac{4}{3} \pi a^3\). As the sphere is uniform, the mass of the shell is
\(
d m=\frac{M}{\frac{4}{3} \pi a^3} 4 \pi x^2 d x=\frac{3 M}{a^3} x^2 d x .
\)
The potential due to this shell at the point \(P\) is \(d V=-\frac{G d m}{r}\) if \(x<r\) and \(d V=-\frac{G d m}{x}\) if \(x>r\)

Case 1: Potential at an external point

Suppose the point \(P\) is outside the sphere (shown in the figure). The potential at \(P\) due to the shell considered is
\(
d V=-\frac{G d m}{r} .
\)
Thus, the potential due to the whole sphere is
\(
\begin{aligned}
V &=\int d V=-\frac{G}{r} \int d m \\
&=-\frac{G M}{r} \dots(1.5)
\end{aligned}
\)
The gravitational potential due to a uniform sphere at an external point is same as that due to a single particle of equal mass placed at its centre.

Case 2: Potential at an internal point

Let us divide the sphere in two parts by imagining a concentric spherical surface passing through \(P\). The inner part has a mass
\(M^{\prime}=\frac{M}{\frac{4}{3} \pi a^3} \times \frac{4}{3} \pi r^3=\frac{M r^3}{a^3} .\)

The potential at \(P\) due to this inner part is by equation (1.5)
\(
\begin{aligned}
V_1 &=-\frac{G M^{\prime}}{r} \\
&=-\frac{G M r^2}{a^3} \dots(i)
\end{aligned}
\)
To get the potential at \(P\) due to the outer part of the sphere, we divide this part in concentric shells. The mass of the shell between radii \(x\) and \(x+d x\) is
\(
d m=\frac{M}{\frac{4}{3} \pi a^3} 4 \pi x^2 d x=\frac{3 M x^2 d x}{a^3} .
\)
The potential at \(P\) due to this shell is,
\(
\frac{-G d m}{x}=-3 \frac{G M}{a^3} x d x .
\)
The potential due to the outer part is
\(
\begin{aligned}
V_2 &=\int_r^a-\frac{3 G M}{a^3} x d x \\
&=-\frac{3 G M}{a^3}\left[\frac{x^2}{2}\right]_r^a \\
&=\frac{-3 G M}{2 a^3}\left(a^2-r^2\right) \dots(ii)
\end{aligned}
\)
By (i) and (ii) the total potential at \(P\) is
\(
\begin{aligned}
V &=V_1+V_2 \\
&=-\frac{G M r^2}{a^3}-\frac{3 G M}{2 a^3}\left(a^2-r^2\right) \\
&=-\frac{G M}{2 a^3}\left(3 a^2-r^2\right) \dots(1.6)
\end{aligned}
\)
At the centre of the sphere the potential is
\(
V=-\frac{3 G M}{2 a} \text {. }
\)

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 kg.

The diagram represents the Earth’s gravitational field. The lines show the direction of the force that acts on a mass that is within the field.

This diagram shows that:

• gravitational forces are always attractive – the Earth cannot repel any objects
• the Earth’s gravitational pull acts towards the centre of the Earth
• the Earth’s gravitational field is radial; the field lines become less concentrated with increasing distance from the Earth.

The force exerted on an object in a gravitational field depends on its position. The less concentrated the field lines, the smaller the force.

We define the intensity of gravitational field \(\vec{E}\) at a point by the equation
\(
\overrightarrow{E}=\frac{\overrightarrow{F}}{m} 
\)
where \(\vec{F}\) is the force exerted by the field on a body of mass \(m\) placed in the field. Quite often the intensity of the gravitational field is abbreviated as the gravitational field. Its SI unit is \(\mathrm{N} \mathrm{kg}^{-1}\).

The total gravitational field at a point is given by the vector sum of the fields due to individual masses.

\(\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}\)

Intensity of the gravitational field near the surface of the earth

If a mass \(m\) is placed close to the surface of the earth, the force on it is \(m g\). We say that the earth has set up a gravitational field and this field exerts a force on the mass. The intensity of the field is
\(
\vec{E}=\frac{\vec{F}}{m}=\frac{m \vec{g}}{m}=\vec{g} .
\)
Thus, the intensity of the gravitational field near the surface of the earth is equal to the acceleration due to gravity.

Example 4: A particle of mass \(50 \mathrm{~g}\) experiences a gravitational force of \(2.0 \mathrm{~N}\) when placed at a particular point. Find the gravitational field at that point.

Solution: The gravitational field has a magnitude
\(
E=\frac{F}{m}=\frac{2.0 \mathrm{~N}}{\left(50 \times 10^{-3} \mathrm{~kg}\right)}=40 \mathrm{~N} \mathrm{~kg}^{-1} \text {. }
\)
This field is along the direction of the force.

Relation Between Gravitational Field and Potential

Suppose the gravitational field at a point \(\vec{r}\) due to a given mass distribution is \(\vec{E}\). By definition, the force on a particle of mass \(m\) when it is at \(\vec{r}\) is
\(
\vec{F}=m \vec{E}
\)
As the particle is displaced from \(\vec{r}\) to \(\vec{r}+d \vec{r}\) the work done by the gravitational force on it is
\(
\begin{aligned}
d W &=\vec{F} \cdot d \vec{r} \\
&=m \vec{E} \cdot d \vec{r}
\end{aligned}
\)
The change in potential energy during this displacement is
\(
d U=-d W=-m \vec{E} \cdot d \vec{r} 
\)
The change in potential is, by equation (1.7),
\(d V=\frac{d U}{m}=-\vec{E} \cdot d \vec{r} \dots(1.7)\)
Integrating between \(\overrightarrow{r_1}\) and \(\overrightarrow{r_2}\)
\(
V\left(\overrightarrow{r_2}\right)-V\left(\overrightarrow{r_1}\right)=-\int_{\overrightarrow{r_1}}^{\vec{r}_2} \vec{E} \cdot d \vec{r} .
\)
If \(\overrightarrow{r_1}\) is taken at the reference point, \(V\left(\overrightarrow{r_1}\right)=0\). The potential \(V(\vec{r})\) at any point \(\vec{r}\) is, therefore,
\(
V(\vec{r})=-\int_{\overrightarrow{r_0}}^{\vec{r}} \vec{E} \cdot d \vec{r}
\)
where \(\overrightarrow{r_0}\) denotes the reference point.
If we work in Cartesian coordinates, we can write
\(
\vec{E}=E_x\hat{i}+E_y\hat{j}+E_z\hat{k}
\)
and \(\quad d \vec{r}=d x\hat{i}+d y\hat{j}+d z\hat{k}\)
so that \(\vec{E} \cdot d \vec{r}=E_x d x+E_y d y+E_z d z\).
Equation (11.14) may be written as
\(
d V=-E_x d x-E_y d y-E_z d z
\)
If \(y\) and \(z\) remain constant, \(d y=d z=0\).
Thus, \(\quad E_x=-\frac{\partial V}{\partial x}\)
Similarly, \(\quad E_y=-\frac{\partial V}{\partial y}\) and \(E_z=-\frac{\partial V}{\partial z}\).
The symbol \(\frac{\partial}{\partial x}\) means partial differentiation with respect to \(x\) treating \(y\) and \(z\) to be constants. Similarly for \(\frac{\partial}{\partial y}\) and \(\frac{\partial}{\partial z}\).

Example 5: The gravitational field due to a mass distribution is given by \(E=K / x^3\) in \(X\)-direction. Taking the gravitational potential to be zero at infinity, find its value at a distance \(x\).

Solution: The potential at a distance \(x\) is
\(
\begin{aligned}
V(x) &=-\int_{\infty}^x E d x=-\int_{\infty}^x \frac{K}{x^3} d x \\
&=\left[\frac{K}{2 x^2}\right]_{\infty}^x=\frac{K}{2 x^2} 
\end{aligned}
\)

Example 6: The gravitational potential due to a mass distribution is \(V=\frac{A}{\sqrt{x^2+a^2}}\). Find the gravitational field.

Solution: \(\quad V=\frac{A}{\sqrt{x^2+a^2}}=A\left(x^2+a^2\right)^{-1 / 2}\)
If the gravitational field is \(E\),
\(
\begin{aligned}
E_x &=-\frac{\partial V}{\partial x}=-A\left(-\frac{1}{2}\right)\left(x^2+a^2\right)^{-3 / 2}(2 x) \\
&=\frac{A x}{\left(x^2+a^2\right)^{3 / 2}} \\
E_y &=-\frac{\partial V}{\partial y}=0 \text { and } E_z=-\frac{\partial V}{\partial z}=0 .
\end{aligned}
\)
The gravitational field is \(\frac{A x}{\left(x^2+a^2\right)^{3 / 2}}\) in the \(x\)-direction.

Calculation of Gravitational Field

(A) Field due to a Point Mass

Suppose a particle of mass \(M\) is placed at a point \(O\) (figure below) and a second particle of mass \(m\) is placed at a point \(P\). Let \(O P=r\).

The mass \(M\) creates a field \(\vec{E}\) at the site of mass \(m\) and this field exerts a force
\(
\vec{F}=m \vec{E}
\)
on the mass \(m\). But the force \(\vec{F}\) on the mass \(m\) due to the mass \(M\) is
\(
F=\frac{G M m}{r^2}
\)
acting along \(\overrightarrow{P O}\). Thus, the gravitational field at \(P\) is
\(
E=\frac{G M}{r^2} \dots(1.8)
\)
along \(\overrightarrow{P O}\). If \(O\) is taken as the origin, the position vector of mass \(m\) is \(\vec{r}=\overrightarrow{O P}\). Equation (1.8) may be rewritten in vector form as
\(
=-\frac{G M}{r^2} \overrightarrow{e_r} \dots(1.9)
\)
where \(\overrightarrow{e_r}\) is the unit vector along \(\vec{r}\).

(B) Field due to a Uniform Circular Ring at a Point on its Axis

The figure below shows a uniform circular ring of radius \(a\) and mass \(M\). Let \(P\) be a point on its axis at a distance \(r\) from the centre. We have to obtain the gravitational field at \(P\) due to the ring. By symmetry, the field must be towards the centre that is along \(\overrightarrow{P O}\)

Consider any particle of mass \(d m\) on the ring, say at point \(A\). The distance of this particle from \(P\) is \(A P=z=\sqrt{a^2+r^2}\). The gravitational field at \(P\) due to \(d m\) is along \(\overrightarrow{P A}\) and its magnitude is
\(
d E=\frac{G d m}{z^2} .
\)
The component along \(P O\) is
\(
d E \cos \alpha=\frac{G d m}{z^2} \cos \alpha .
\)
The net gravitational field at \(P\) due to the ring is
\(
\begin{aligned}
E &=\int \frac{G d m}{z^2} \cos \alpha=\frac{G \cos \alpha}{z^2} \int d m=\frac{G M \cos \alpha}{z^2} \\
&=\frac{G M r}{\left(a^2+r^2\right)^{3 / 2}} \dots(1.10)
\end{aligned}
\)
The field is directed towards the centre of the ring.

(C) Field due to a Uniform Disc at a Point on its Axis

The situation is shown in figure below. Let the mass of the disc be \(M\) and its radius be \(a\). Let \(O\) be the centre of the disc and \(P\) be a point on its axis at a distance \(r\) from the centre. We have to find the gravitational field at \(P\) due to the disc.

Let us draw a circle of radius \(x\) with the centre at O. We draw another concentric circle of radius \(x+d x\). The part of the disc enclosed between these two circles can be treated as a uniform ring of radius \(x\). The point \(P\) is on its axis at a distance \(r\) from the centre. The area of this ring is \(2 \pi x d x\). The area of the whole disc is \(\pi a^2\). As the disc is uniform, the mass of this ring is
\(
\begin{aligned}
d m &=\frac{M}{\pi a^2} 2 \pi x d x \\
&=\frac{2 M x d x}{a^2} .
\end{aligned}
\)
The gravitational field at \(P\) due to the ring is, by equation (1.10),
\(
\begin{aligned}
d E &=\frac{G\left(\frac{2 M x d x}{a^2}\right) r}{\left(r^2+x^2\right)^{3 / 2}} \\
&=\frac{2 G M r}{a^2} \frac{x d x}{\left(r^2+x^2\right)^{3 / 2}} .
\end{aligned}
\)
As \(x\) varies from 0 to \(a\), the rings cover up the whole disc. The field due to each of these rings is in the same direction \(P O\). Thus, the net field due to the whole disc is along \(P O\) and its magnitude is
\(
\begin{aligned}
E &=\int_0^a \frac{2 G M r}{a^2} \frac{x d x}{\left(r^2+x^2\right)^{3 / 2}} \\
&=\frac{2 G M r}{a^2} \int_0^a \frac{x d x}{\left(r^2+x^2\right)^{3 / 2}} \dots(i)
\end{aligned}
\)
Let \(r^2+x^2=z^2\).
Then \(2 x d x=2 z d z\) and
\(
\begin{aligned}
& \int \frac{x d x}{\left(r^2+x^2\right)^{3 / 2}}=\int \frac{z d z}{z^3} \\
=& \int \frac{1}{z^2} d z=-\frac{1}{z}=-\frac{1}{\sqrt{r^2+x^2}} .
\end{aligned}
\)
From (i), \(\quad E=\frac{2 G M r}{a^2}\left[-\frac{1}{\sqrt{r^2+x^2}}\right]_0^a\)
\(
=\frac{2 G M r}{a^2}\left[\frac{1}{r}-\frac{1}{\sqrt{r^2+a^2}}\right] \dots(1.11)
\)
Equation (1.11) may be expressed in terms of the angle \(\theta\) subtended by a radius of the disc at \(P\) as,
\(
E=\frac{2 G M}{a^2}(1-\cos \theta)
\)

(D) Field due to a Uniform Thin Spherical Shell

The shaded ring shown in the figure below has mass \(d m=\frac{M}{2} \sin \theta d \theta\). The field at \(P\) due to this ring is
\(
d E=\frac{G d m}{z^2} \cos \alpha=\frac{G M}{2} \frac{\sin \theta d \theta \cos \alpha}{z^2} \dots(i)
\)

From the triangle \(O A P\),
\(z^2=a^2+r^2-2 a r \cos \theta\)
or, \(\quad 2 z d z=2 a r \sin \theta d \theta\)
or, \(\sin \theta d \theta=\frac{z d z}{a r} \dots(ii)\)
\(a^2=z^2+r^2-2 z r \cos \alpha\)
or, \(\quad \cos \alpha=\frac{z^2+r^2-a^2}{2 z r} \dots(iii)\) 
Putting from (ii) and (iii) in (i),
\(
d E=\frac{G M}{4 a r^2}\left(1-\frac{a^2-r^2}{z^2}\right) d z
\)
or, \(\quad \int d E=\frac{G M}{4 a r^2}\left[z+\frac{a^2-r^2}{z}\right]\)

Case 1: \(P\) is outside the shell \((r>a)\)

In this case \(z\) varies from \(r-a\) to \(r+a\). The field due to the whole shell is
\(
E=\frac{G M}{4 a r^2}\left[z+\frac{a^2-r^2}{z}\right]_{r-a}^{r+a}=\frac{G M}{r^2} .
\)
We see that the shell may be treated as a point particle of the same mass placed at its centre to calculate the gravitational field at an external point.

Case 2: \(P\) is inside the shell \((r<a)\)

In this case \(z\) varies from \(a-r\) to \(a+r\) (figure below).

The field at \(P\) due to the whole shell is
\(
E=\frac{G M}{4 a r^2}\left[z+\frac{a^2-r^2}{z}\right]_{a-r}^{a+r}=0 .
\)
Hence the field inside a uniform spherical shell is zero.

(E) Gravitational Field due to a Uniform Solid Sphere

Case 1: Field at an external point

Let the mass of the sphere be \(M\) and its radius be \(a\). We have to calculate the gravitational field due to the sphere at a point outside the sphere at a distance \(r\) from the centre. The figure below shows the situation. The centre of the sphere is at \(O\) and the field is to be calculated at \(P\).

Let us divide the sphere into thin spherical shells each centred at \(O\). Let the mass of one such shell be \(d m\). To calculate the gravitational field at \(P\), we can replace the shell by a single particle of mass \(d m\) placed at the centre of the shell that is at \(O\). The field at \(P\) due to this shell is then
\(
d E=\frac{G d m}{r^2}
\)
towards \(P O\). The field due to the whole sphere may be obtained by summing the fields of all the shells making the solid sphere.
Thus, \(\quad E=\int d E\)
\(
\begin{aligned}
&=\int \frac{G d m}{r^2}=\frac{G}{r^2} \int d m \\
&=\frac{G M}{r^2} \dots(1.12)
\end{aligned}
\)
Thus, a uniform sphere may be treated as a single particle of equal mass placed at its centre for calculating the gravitational field at an external point.
This allows us to treat the earth as a point particle placed at its centre while calculating the force between the earth and an apple.

Case 2: Field at an internal point

Suppose the point \(P\) is inside the solid sphere (figure below). In this case \(r<a\). The sphere may be divided into thin spherical shells all centered at \(O\). Suppose the mass of such a shell is \(d m\). If the radius of the shell is less than \(r\), the point \(P\) is outside the shell and the field due to the shell is
\(
d E=\frac{G d m}{r^2} \text { along } P O
\)

If the radius of the shell considered is greater than \(r\), the point \(P\) is internal and the field due to such a shell is zero. The total field due to the whole sphere is obtained by summing the fields due to all the shells. As all these fields are along the same direction, the net field is
\(
\begin{aligned}
E &=\int d E \\
&=\int \frac{G d m}{r^2}=\frac{G}{r^2} \int d m \dots(i)
\end{aligned}
\)
Only the masses of the shells with radii less than \(r\) should be added to get \(\int d m\). These shells form a solid sphere of radius \(r\). The volume of this sphere is \(\frac{4}{3} \pi r^3\). The volume of the whole sphere is \(\frac{4}{3} \pi a^3\). As the given sphere is uniform, the mass of the sphere of radius \(r\) is
\(
\frac{M}{\frac{4}{3} \pi a^3} \cdot\left(\frac{4}{3} \pi r^3\right)=\frac{M r^3}{a^3} .
\)
Thus, \(\quad \int d m=\frac{M r^3}{a^3}\)
and by (i) \(E=\frac{G}{r^2} \frac{M r^3}{a^3}\)
\(
=\frac{G M}{a^3} r \text {. } \dots(1.13)
\)
The gravitational field due to a uniform sphere at an internal point is proportional to the distance of the point from the centre of the sphere. At the centre itself, \(r=0\) and the field is zero. This is also expected from symmetry because any particle at the centre is equally pulled from all sides and the resultant must be zero.

At the surface of the sphere, \(r=a\) and
\(
E=\frac{G M}{a^2} .
\)
The formula (1.12) for the field at an external point also gives \(E=\frac{G M}{a^2}\) at the surface of the sphere. The two formulae agree at \(r=a\). The figure below shows graphically the variation of gravitational field due to a solid sphere with the distance from its centre.

Example 7: Find the gravitational field due to the moon at its surface. The mass of the moon is \(7.36 \times 10^{22} \mathrm{~kg}\) and the radius of the moon is \(1.74 \times 10^6 \mathrm{~m}\). Assume the moon to be a spherically symmetric body.

Solution: To calculate the gravitational field at an external point, the moon may be replaced by a single particle of equal mass placed at its centre. Then the field at the surface is
\(
\begin{aligned}
E &=\frac{G M}{a^2} \\
&=\frac{6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2 \times 7.36 \times 10^{22} \mathrm{~kg}}{\left(1.74 \times 10^6 \mathrm{~m}\right)^2} \\
&=1.62 \mathrm{~N} \mathrm{~kg}^{-1}
\end{aligned}
\)
This is about one-sixth of the gravitational field due to the earth at its surface.

Variation in the Value of \(g\)

The acceleration due to gravity is given by
\(
g=\frac{F}{m}
\)
where \(F\) is the force exerted by the earth on an object of mass \(m\). This force is affected by a number of factors and hence \(g\) also depends on these factors.

(a) Height from the Surface of the Earth

If the object is placed at a distance \(h\) above the surface of the earth, the force of gravitation on it due to the earth is
\(
F=\frac{G M m}{(R+h)^2}
\)
where \(M\) is the mass of the earth and \(R\) is its radius. Thus, \(\quad g=\frac{F}{m}=\frac{G M}{(R+h)^2}\).
We see that the value of \(g\) decreases as one goes up. We can write,
\(
g=\frac{G M}{R^2\left(1+\frac{h}{R}\right)^2}=\frac{g_0}{\left(1+\frac{h}{R}\right)^2}
\)
where \(g_0=\frac{G M}{R^2}\) is the value of \(g\) at the surface of the earth. If \(\text { If } h \ll R \text {, }\)
\(
g=g_0\left(1+\frac{h}{R}\right)^{-2} \approx g_0\left(1-\frac{2 h}{R}\right) .
\)
If one goes a distance \(h\) inside the earth such as in mines, the value of \(g\) again decreases. The force by the earth is, by equation (1.13),
\(
\begin{aligned}
F &=\frac{G M m}{R^3}(R-h) \\
g &=\frac{F}{m}=\frac{G M}{R^2}\left(\frac{R-h}{R}\right) \\
&=g_0\left(1-\frac{h}{R}\right) .
\end{aligned}
\)
The value of \(g\) is maximum at the surface of the earth and decreases with the increase in height as well as with depth.

Example 8: Calculate the value of acceleration due to gravity at a point (a) \(5.0 \mathrm{~km}\) above the earth’s surface and (b) \(5.0 \mathrm{~km}\) below the earth’s surface. radius of earth \(=6400 \mathrm{~km}\) and the value of \(g\) at the surface of the earth is \(9.80 \mathrm{~m} \mathrm{~s}^{-2}\).

Solution:

(a) The value of \(g\) at a height \(h\) is (for \(h \ll R\) )
\(
\begin{aligned}
g &=g_0\left(1-\frac{2 h}{R}\right) \\
&=\left(9.80 \mathrm{~m} \mathrm{~s}^{-2}\right)\left(1-\frac{2 \times 5.0 \mathrm{~km}}{6400 \mathrm{~km}}\right) \\
&=9.78 \mathrm{~m} \mathrm{~s}^{-2} .
\end{aligned}
\)
(b) The value at a depth \(h\) is
\(
\begin{aligned}
g &=g_0\left(1-\frac{h}{R}\right) \\
&=\left(9.8 \mathrm{~m} \mathrm{~s}^{-2}\right)\left(1-\frac{5 \cdot 0 \mathrm{~km}}{6400 \mathrm{~km}}\right) \\
&=9.79 \mathrm{~m} \mathrm{~s}^{-2} .
\end{aligned}
\)

(b) Rotation of the Earth

As the earth rotates about its own axis the frame attached to the earth is noninertial. If we wish to use the familiar Newton’s laws, we have to include pseudo forces. For an object at rest with respect to the earth, a centrifugal force \(m \omega^2 r\) is to be added where \(m\) is the mass of the object, \(\omega\) is the angular velocity of the earth and \(r\) is the radius of the circle in which the particle rotates.

If the colatitude of the location of the particle is \(\theta\) (figure below),r=R \(\sin \theta\) where \(R\) is the radius of the earth. Acceleration of an object falling near the earth’s surface, as measured from the earth frame, is \(F / m\) where \(F\) is the vector sum of the gravitational force \(\frac{G M m}{R^2}=m g\) and the centrifugal force \(m \omega^2 r=\) \(m \omega^2 R \sin \theta\). The acceleration \(F / m=g^{\prime}\) is the apparent value of the acceleration due to gravity.

At the equator, \(\theta=\pi / 2\) and the centrifugal force is just opposite to the force of gravity. The resultant of these two is

\(
\begin{aligned}
& F=m g-m \omega^2 R \\
\text { or, } & g^{\prime}=g-\omega^2 R .
\end{aligned}
\)

At the poles, \(\theta=0\) and the centrifugal force \(m \omega{ }^2 R \sin \theta=0\). Thus, \(F=m g\) and \(g^{\prime}=g\). Thus, the observed value of the acceleration due to gravity is minimum at the equator and is maximum at the poles. 

(c) Nonsphericity of the Earth

All formulae and equations have been derived by assuming that the earth is a uniform solid sphere. The shape of the earth slightly deviates from the perfect sphere. The radius in the equatorial plane is about \(21 \mathrm{~km}\) larger than the radius along the poles. Due to this, the force of gravity is more at the poles and less at the equator. The value of \(g\) is accordingly larger at the poles and less at the equator. Note that due to rotation of earth also, the value of \(g\) is smaller at the equator than that at the poles.

(d) Nonuniformity of the Earth

The earth is not a uniformly dense object. There are a variety of minerals, metals, water, oil, etc., inside the earth. Then at the surface there are mountains, seas, etc. Due to these nonuniformities in the mass distribution, the value of \(g\) is locally affected.

“Weighing” the Earth

The force exerted by the earth on a body is called the weight of the body. In this sense “weight of the earth” is a meaningless concept. However, the mass of the earth can be determined by noting the acceleration due to gravity near the surface of the earth. We have,
\(g=\frac{G M}{R^2}\)
or, \(\quad M=g R^2 / G\)
Putting \(g=9.8 \mathrm{~m} \mathrm{~s}^{-2}, R=6400 \mathrm{~km}\)
and
\(
G=6.67 \times 10^{-11} \frac{\mathrm{N}^{-\mathrm{m}^2}}{\mathrm{~kg}^2}
\)
the mass of the earth comes out to be \(5 \cdot 98 \times 10^{24} \mathrm{~kg}\).

Planets and Satellites

Planets move round the sun due to the gravitational attraction of the sun. The path of these planets are elliptical with the sun at a focus. However, the difference in major and minor axes is not large. The orbits can be treated as nearly circular for not-too-sophisticated calculations. Let us derive certain characteristics of the planetary motion in terms of the radius of the orbit assuming it to be perfectly circular.

Let the mass of the sun be \(M\) and that of the planet under study be \(m\). The mass of the sun is many times larger than the mass of the planet. The sun may, therefore, be treated as an inertial frame of reference.

Speed of the Planet (\(v\)) in the Orbit

Let the radius of the orbit be \(a\) and the speed of the planet in the orbit be \(v\). By Newton’s second law, the force on the planet equals its mass times the acceleration. Thus,
\(\frac{G M m}{a^2}=m\left(\frac{v^2}{a}\right)\)
or, \(\quad v=\sqrt{\frac{G M}{a}} \dots(1.14)\) 
The speed of a planet is inversely proportional to the square root of the radius of its orbit.

Time Period \(T\)

The time taken by a planet in completing one revolution is its time period \(T\). In one revolution it covers a linear distance of \(2 \pi a\) at speed \(v\). Thus,
\(
\begin{aligned}
T &=\frac{2 \pi a}{v} \\
&=\frac{2 \pi a}{\sqrt{\frac{G M}{a}}}=\frac{2 \pi}{\sqrt{G M}} a^{3 / 2}
\end{aligned}
\)
or, \(\quad T^2=\frac{4 \pi^2}{G M} a^3 \dots(1.15)\)

Energy

The kinetic energy of the planet is
\(
K=\frac{1}{2} m v^2 \text {. }
\)
Using (1.14),
\(
K=\frac{1}{2} m \frac{G M}{a}=\frac{G M m}{2 a} .
\)
The gravitational potential energy of the sun-planet system is
\(
U=-\frac{G M m}{a} .
\)
The total mechanical energy of the sun-planet system is
\(
E=K+U=\frac{G M m}{2 a}-\frac{G M m}{a}=-\frac{G M m}{2 a} .
\)
The total energy is negative. This is true for any bound system if the potential energy is taken to be zero at infinite separation.

Satellite

Satellites are launched from the earth so as to move around it. A number of rockets are fired from the satellite at the proper time to establish the satellite in the desired orbit. Once the satellite is placed in the desired orbit with the correct speed for that orbit, it will continue to move in that orbit under the gravitational attraction of the earth. All the equations derived above for planets are also true for satellites with \(M\) representing the mass of the earth and \(m\) representing the mass of the satellite.

Example 9: A satellite is revolving round the earth at a height of \(600 \mathrm{~km}\). Find (a) the speed of the satellite and (b) the time period of the satellite. Radius of the earth \(=6400 \mathrm{~km}\) and mass of the earth \(=6 \times 10^{24} \mathrm{~kg}\).

Solution: The distance of the satellite from the centre of the earth is \(6400 \mathrm{~km}+600 \mathrm{~km}=7000 \mathrm{~km}\).
The speed of the satellite is
\(
\begin{aligned}
v &=\sqrt{\frac{G M}{a}} \\
&=\sqrt{\frac{6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2 \times 6 \times 10^{24} \mathrm{~kg}}{7000 \times 10^3 \mathrm{~m}}} \\
&=7.6 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}=7.6 \mathrm{~km} \mathrm{~s}^{-1} 
\end{aligned}
\)
The time period is
\(
\begin{aligned}
T &=\frac{2 \pi a}{v} \\
&=\frac{2 \pi \times 7000 \times 10^3 \mathrm{~m}}{7^{.} 6 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}}=5.8 \times 10^3 \mathrm{~s} 
\end{aligned}
\)

Geostationary Satellite

The earth rotates about its own axis (the line joining the north pole and the south pole) once in 24 hours. Suppose a satellite is established in an orbit in the plane of the equator. Suppose the height is such that the time period of the satellite is 24 hours and it moves in the same sense as the earth. The satellite will always be overhead a particular place on the equator. As seen from the earth, this satellite will appear to be stationary. Such a satellite is called a geostationary satellite. Such satellites are used for telecommunication, weather forecasting and other applications.
According to equation (1.15),
\(
\begin{aligned}
T^2 &=\frac{4 \pi^2}{G M} a^3 \\
a &=\left(\frac{G M T^2}{4 \pi^2}\right)^{1 / 3} 
\end{aligned}
\)
Putting the values of \(G, M=\left(6 \times 10^{24} \mathrm{~kg}\right)\) and \(T=(24\) hours \()\); the radius of the geostationary orbit comes out to be \(a=4.2 \times 10{ }^4 \mathrm{~km}\). The height above the surface of the earth is about \(3.6 \times 10^4 \mathrm{~km}\).

Weightlessness Of a Satellite

A satellite moves round the earth in a circular orbit under the action of gravity. The acceleration of the satellite is \(\frac{G M}{R^2}\) towards the centre of the earth, where \(M\) is the mass of the earth and \(R\) is the radius of the orbit of the satellite. Consider a body of mass \(m\) placed on a surface inside a satellite moving round the earth. The forces on the body are
(a) the gravitational pull of the earth \(=\frac{G M m}{R^2}\),
(b) the contact force \(\mathcal{N}\) by the surface.
By Newton’s law,
\(
G \frac{M m}{R^2}-\mathcal{N}=m\left(\frac{G M}{R^2}\right) \text { or, } \mathcal{N}=0 .
\)
Thus, the surface does not exert any force on the body and hence its apparent weight is zero. No support is needed to hold a body in the satellite.

One can analyse the situation from the frame of the satellite. Working in the satellite frame we have to add a centrifugal force on all bodies. If the mass of a body is \(m\), the centrifugal force is \(m\left(\frac{G M}{R^2}\right)\) away from the centre of the earth. This pseudo force exactly balances the weight of the body which is \(\frac{G M m}{R^2}\) towards the centre of the earth. A body needs no support to stay at rest in the satellite and hence all positions are equally comfortable. Water will not fall down from the glass even if it is inverted. It will act like a “gravity-free hall”. Such a state is called weightlessness.

It should be clear that the earth still attracts a body with the same force \(\frac{G M m}{R^2}\). The feeling of weightlessness arises because one stays in a rotating frame.