Gravitation

Gravitational Potential

At a point in a gravitational field potential \(V\) is defined as negative of work done per unit mass in shifting a test mass from some reference point (usually at infinity) to the given point i.e.,

\(V=-\frac{W}{m}=-\int{\frac{\overrightarrow{F}.d\overrightarrow{r\,}}{m}}\) \(=-\int{\overrightarrow{I\,}.d\overrightarrow{r\,}}\)     [As \(\frac{F}{m}=I\)]

\(\therefore h=0\)

i.e., negative gradient of potential gives intensity of field or potential is a scalar function of position whose space derivative gives intensity. Negative sign indicates that the direction of intensity is in the direction where the potential decreases.

(i) It is a scalar quantity because it is defined as work done per unit mass.

(ii) Unit : Joule/kg or m²/sec²

(iii)Dimension : [M⁰L²T^–2]

(iv)If the field is produced by a point mass then \(V=-\int_{{}}^{{}}{I\ dr}\)\(T=2\pi \sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \sqrt{\frac{{{R}^{3}}}{g{{R}^{2}}}}=2\pi \sqrt{\frac{R}{g}}\)             [As \(I=-\frac{GM}{{{r}^{2}}}\)]

\(\therefore V=-\frac{GM}{r}+c\)           [Here c = constant of integration]

Assuming reference point at \(\infty\) and potential to be zero there we get   \(0=-\frac{GM}{\infty }+c\,\Rightarrow c=0\)

\(\therefore\)  Gravitational potential \(V=-\frac{GM}{r}\)

(v) Gravitational potential difference : It is defined as the work done to move a unit mass from one point to the other in the gravitational field. The gravitational potential difference in bringing unit test mass m from point A to point B under the gravitational influence of source mass M is \(r\) 

(vi) Potential due to large numbers of particle is given by scalar addition of all the potentials

\(V={{V}_{1}}+{{V}_{2}}+{{V}_{3}}+..........\)

\(=-\frac{GM}{{{r}_{1}}}-\frac{GM}{{{r}_{2}}}-\frac{GM}{{{r}_{3}}}........\)

\(=-G\sum\limits_{i=1}^{i=n}{\frac{{{M}_{i}}}{{{r}_{i}}}}\)

Gravitational Potential for Different Bodies

(1) Potential due to uniform ring

(2) Potential due to spherical shell

(3) Potential due to uniform solid sphere

Gravitational Potential Energy

The gravitational potential energy of a body at a point is defined as the amount of work done in bringing the body from infinity to that point against the gravitational force.

\(W=\int_{\infty }^{r}{\frac{GMm}{{{x}^{2}}}dx=-GMm\,\left[ \frac{1}{x} \right]_{\infty }^{r}}\)

\(W=-\frac{GMm}{r}\)

This work done is stored inside the body as its gravitational potential energy

\(\therefore U=-\frac{GMm}{r}\)

(i) Potential energy is a scalar quantity.

(ii) Unit : Joule

(iii) Dimension : [ML²T^–2]

(iv) Gravitational potential energy is always negative in the gravitational field because the force is always attractive in nature.

(v) As the distance \(r\) increases, the gravitational potential energy becomes less negative i.e., it increases.

(vi) If \(r=\infty\) then it becomes zero (maximum)

(vii) In case of discrete distribution of masses

Gravitational potential energy

\(U=\sum {{u}_{i}}=-\left[ \frac{G{{m}_{1}}{{m}_{2}}}{{{r}_{12}}}+\frac{G{{m}_{2}}{{m}_{3}}}{{{r}_{23}}}+........ \right]\)

(viii) If the body of mass $m$ is moved from a point at a distance \({{r}_{1}}\) to a point at distance \({{r}_{2}}({{r}_{1}}>{{r}_{2}})\) then change in potential energy \(\Delta U=\int_{{{r}_{1}}}^{{{r}_{2}}}{\frac{GMm}{{{x}^{2}}}}dx=-GMm\,\left[ \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right]\)

or \(\Delta U=GMm\,\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\)

As \({{r}_{1}}\)is greater than \(T=2\pi \,\,\sqrt{\frac{{{\left( R+h \right)}^{3}}}{g\,{{R}^{2}}}}\), the change in potential energy of the body will be negative. It means that if a body is brought closer to earth it's potential energy decreases.

(ix) Relation between gravitational potential energy and potential \(U=-\frac{GMm}{r}=m\left[ \frac{-GM}{r} \right]\)

\(=2\,\pi \,\,\sqrt{\frac{R}{g}}{{\left( 1+\frac{h}{R} \right)}^{3/2}}\)

(x) Gravitational potential energy at the centre of earth relative to infinity.

\({{U}_{centre}}=m\,{{V}_{centre}}=m\left( -\frac{3}{2}\frac{GM}{R} \right)=-\frac{3}{2}\frac{GMm}{R}\)

(xi) Gravitational potential energy of a body at height h from the earth surface is given by

\(GM=g{{R}^{2}}\)

Work Done Against Gravity

If the body of mass \(m\) is moved from the surface of earth to a point at distance $h$ above the surface of earth, then change in potential energy or work done against gravity will be

\(T=2\pi \,\sqrt{\frac{{{r}^{3}}}{GM}}=2\pi \,\,\sqrt{\frac{{{r}^{3}}}{g{{R}^{2}}}}\)    

\(W=GMm\left[ \frac{1}{R}-\frac{1}{R+h} \right]\,\, \,\,[As \,\,{{r}_{1}}=R\,\, and \,\,v=\sqrt{\frac{GM}{r}}]\)

\(\Rightarrow W=\frac{GMmh}{{{R}^{2}}\left( 1+\frac{h}{R} \right)}=\frac{mgh}{1+\frac{h}{R}}\)           [As \(\frac{GM}{{{R}^{2}}}=g\)]

 (i) When the distance \(h\) is not negligible and is comparable to radius of the earth, then we will use above formula.

(ii) If \(h=nR\) then \(W=mgR\left( \frac{n}{n+1} \right)\)

(iii) If \(h=R\) then \(W=\frac{1}{2}mgR\)

(iv) If \(h\)is very small as compared to radius of the earth then term \(h/R\) can be neglected

From \(W=\frac{mgh}{1+h/R}=mgh\)    \(\left[ \text{As }\,\,\frac{h}{R}\to 0 \right]\)

Escape Velocity

The minimum velocity with which a body must be projected up so as to enable it to just overcome the gravitational pull, is known as escape velocity.

The work done to displace a body from the surface of earth (r = R) to infinity (\(r=\infty\)) is

\(g=\frac{4}{3}\pi \rho GR=-GMm\,\left[ \frac{1}{\infty }-\frac{1}{R} \right]\)

\(\Rightarrow W=\frac{GMm}{R}\)

This work required to project the body so as to escape the gravitational pull is performed on the body by providing an equal amount of kinetic energy to it at the surface of the earth.

If \({{v}_{e}}\) is the required escape velocity, then kinetic energy which should be given to the body is \(\frac{1}{2}mv_{e}^{2}\)

\(\Rightarrow{{v}_{e}}=\sqrt{2gR}\)             [As \(=-\frac{3}{2}\frac{GMm}{R}\)]s

or \({{v}_{e}}=\sqrt{2\times \frac{4}{3}\pi \rho GR\times R}\Rightarrow {{v}_{e}}=R\sqrt{\frac{8}{3}\pi G\rho }\)        [As \(g=\frac{4}{3}\pi \rho GR\)]

(i) Escape velocity is independent of the mass and direction of projection of the body.

(ii) Escape velocity depends on the reference body. Greater the value of \((M/R)\) or \(v\propto \frac{1}{\sqrt{{{r}^{n}}-1}}\) for a planet, greater will be escape velocity.

(iii) For the earth as \(g=9.8m/{{s}^{2}}\) and \(R=6400\,km\)

\({{v}_{escape}}=\sqrt{2}\,\,\,{{v}_{orbital}}\)

(iv) A planet will have atmosphere if the velocity of molecule in its atmosphere \(\left[ {{v}_{rms}}=\sqrt{\frac{3RT}{M}} \right]\) is lesser than escape velocity. This is why earth has atmosphere (as at earth \(v=\frac{{{v}_{e}}}{\sqrt{2}}\)) while moon has no atmosphere (as at moon \({{v}_{rms}}>{{v}_{e}}\))

(v) If a body projected with velocity lesser than escape velocity (v < v_e), it will reach a certain maximum height and then may either move in an orbit around the planet or may fall down back to the planet.

(vi) Maximum height attained by body : Let a projection velocity of body (mass \(m\)) is \(v\), so that it attains a maximum height \(h\). At maximum height, the velocity of particle is zero, so kinetic energy is zero.

By the law of conservation of energy

Total energy at surface = Total energy at height \(h\).

\(\Rightarrow \ -\frac{GMm}{R}+\frac{1}{2}m{{v}^{2}}=-\frac{GMm}{R+h}+0\)

\(\Rightarrow \frac{{{v}^{2}}}{2}=GM\,\left[ \frac{1}{R}-\frac{1}{R+h} \right]=\frac{GMh}{R(R+h)}\)

\(\Rightarrow \frac{2GM}{{{v}^{2}}R}=\frac{R+h}{h}=1+\frac{R}{h}\)

\(\Rightarrow h=\frac{R}{\left( \frac{2GM}{{{v}^{2}}R}-1 \right)}=\frac{R}{\frac{v_{e}^{2}}{{{v}^{2}}}-1}=R\,\,\left[ \frac{{{v}^{2}}}{v_{e}^{2}-{{v}^{2}}} \right]\)                     

[As\({{v}_{e}}=\sqrt{\frac{2GM}{R}}\,\,\therefore \frac{2GM}{R}=v_{e}^{2}\)]

(vii) If a body is projected with velocity greater than escape velocity \((v>{{v}_{e}})\) then by conservation of energy.

Total energy at surface = Total energy at infinite

\(\frac{1}{2}m{{v}^{2}}-\frac{GMm}{R}=\frac{1}{2}m{{({v}')}^{2}}+0\)

i.e.,\({{({v}')}^{2}}={{v}^{2}}-\frac{2GM}{R}\Rightarrow v{{'}^{2}}={{v}^{2}}-v_{e}^{2}\)      [As \(\frac{2GM}{R}=v_{e}^{2}\)]

\(\therefore {v}'=\sqrt{{{v}^{2}}-v_{e}^{2}}\)

i.e, the body will move in interplanetary or inter stellar space with velocity \(\sqrt{{{v}^{2}}-v_{e}^{2}}\).

(viii) Energy to be given to a stationary object on the surface of earth so that its total energy becomes zero, is called escape energy.

Total energy at the surface of the earth \(=KE+PE=0-\frac{GMm}{R}\)

\(\therefore\) Escape energy \(=\frac{GMm}{R}\)

(ix) If the escape velocity of a body is equal to the velocity of light then from such bodies nothing can escape, not even light. Such bodies are called black holes.

The radius of a black hole is given as

\(R=\frac{2GM}{{{C}^{2}}}\)

[As \(C=\sqrt{\frac{2GM}{R}}\), where \(C\) is the velocity of light]