Gravitational Field, Potential and Velocity of Escape from Earth

Gravitational Field

The space around a body at every point of which another body experiences gravitational field. Limit of a gravitational field is 0 to infinity.

Gravitational field intensity

The gravitational field intensity at any point inside the gravitational field is defined as the gravitational force experienced by a body or unit mass.

If \( F=\frac{GMm}{r^{2}}\) is the gravitational force on a particle of mass m at a point at distance r from the body of mass M inside its gravitational field at that point is given by, \(E=\frac{GM}{r^{2}}\).

Gravitational field intensity at a point is also defined as the space rate of change of gravitational potential i.e. the negative potential gradient i.e.

\(E=-\frac{\mathrm{d} V}{\mathrm{d} x}\) (where dV is the small change in gravitational potential for small change in distance dr.)

Gravitational Potential(V)

The gravitational potential at any point in the gravitational field of a body at any point in the gravitational field of a body is defined as the amount of work done in bringing a unit mass from infinity to that point.

\begin{align*}V=\int_{\infty}^{ r}Edr \end{align*}\begin{align*}V=\int_{\infty}^{ r}\frac{GM}{r^{2}}dr \end{align*}\begin{align*}V=-\frac{GM}{r} \end{align*}

Gravitational Potential Energy(U)

The gravitational potential energy at any point inside the gravitational field of a body is defined as the work done in bringing a body from infinity to that point.

If V be the gravitational potential at a point inside the gravitational field of the body then the gravitational potential energy of a body of mass m at that point is U=mV=\(-\frac{GMm}{r}\).

Velocity of escape:

The velocity required to break free from the gravitational attraction of a massive body without spending further energy.

If the amount of workdone in order to throw the body from the surface of Earth is

\begin{align*}W=\int_{R}^{\infty}Fdx \end{align*}\begin{align*}dW=Fdx \end{align*}\begin{align*}=\int_{R}^{\infty}\frac{GMm}{x^{2}}dx \end{align*}\begin{align*}=-GMm[x^{-1}]_{R}^{\infty} \end{align*}\begin{align*}-GMm[\frac{1}{\infty}-\frac{1}{R}] \end{align*}\begin{align*}\frac{GMm}{R} \end{align*}

If Ve be the velocity of escape from the Earth, the initial K.E. of the body is

\begin{align*}K.E.=1/2mVe^{2} \end{align*}

This K.E. must be equal to the work done in escaping away from Earth.

\begin{align*}1/2mVe^{2}=\frac{GMm}{R} \end{align*}\begin{align*}Ve=\sqrt{\frac{GMm}{R}} \end{align*}\begin{align*}=\sqrt{2gR} \end{align*}\begin{align*}[\because g=\frac{GM}{R^{2}}] \end{align*}

For Earth,

\begin{align*}g=9.8m/s^{2} \end{align*}\begin{align*}R=6400KM=64\times10^{5}m \end{align*}\begin{align*}Ve=\sqrt{2gR} \end{align*}\begin{align*}=11.2Km/sec\end{align*}

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.