JEE Physics Gravitation Notes Set 01

Newton's Law of Gravitation

Any two point masses attract each other along the line joining them. The pull is proportional to the product of the masses and inversely proportional to the square of the separation:
\[ F = \frac{G m_1 m_2}{r^2},\qquad G = 6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} \]
G is a universal constant of nature. The two forces form an action-reaction pair, so they are equal and opposite whatever the masses are. Gravitation is always attractive, acts through empty space and does not depend on the medium in between.

For several masses, the force on one of them is the vector sum of the pulls due to each of the others (principle of superposition). Symmetric arrangements often cancel: a mass at the centroid of an equilateral triangle of equal masses, or at the centre of a regular polygon of equal masses, feels zero net force.

Gravitational Field and Potential

The gravitational field (intensity) at a point is the force on a unit test mass placed there: \( \vec I = \vec F/m \), with unit N/kg. For a point mass M: \( I = GM/r^2 \), directed towards M.
Field of a uniform solid sphere (mass M, radius R):
Outside, \( r \ge R \): \( I = \frac{GM}{r^2} \)
At the surface: \( I = \frac{GM}{R^2} \)
Inside, \( r \lt R \): \( I = \frac{GM r}{R^3} \), growing linearly from zero at the centre
For a thin hollow shell the field is zero everywhere inside and \( GM/r^2 \) outside.

Gravitational potential at a point is the work done per unit mass in bringing a test mass from infinity to that point: \( V = -\frac{GM}{r} \) for a point mass. Potential is a scalar, so potentials of several masses add algebraically. The field is the negative gradient of potential: \( I = -dV/dr \).
Potential of a solid sphere: outside \( V = -\frac{GM}{r} \); inside \( V = -\frac{GM}{2R^3}(3R^2 - r^2) \), which equals \( -\frac{3GM}{2R} \) at the centre. For a hollow shell, potential inside is constant, equal to \( -\frac{GM}{R} \).

Acceleration Due to Gravity

At the surface of the earth, \( g = \frac{GM}{R^2} = \frac43\pi G\rho R \). The mass of the earth is found from \( M = gR^2/G \), and the mean density from \( \rho = \frac{3g}{4\pi GR} \).

Variation of g:
With height h, exact form: \( g_h = g\left(\frac{R}{R+h}\right)^2 \). For \( h \ll R \): \( g_h \approx g\left(1-\frac{2h}{R}\right) \).
With depth d (uniform earth): \( g_d = g\left(1-\frac dR\right) \), reaching zero at the centre.
With latitude \( \lambda \) (rotation): \( g_\lambda = g - \omega^2R\cos^2\lambda \). It is minimum at the equator (\( g-\omega^2R \)) and maximum at the poles (no change).
If the earth spun fast enough that \( \omega = \sqrt{g/R} \) (about \( 1.25\times10^{-3} \) rad/s), bodies at the equator would be weightless.

Mass is constant everywhere, but weight changes with g. A spring balance shows weight, whereas a beam balance compares masses and gives the same reading everywhere.

Worked example 1. At what height above the earth's surface is g equal to one-fourth of its surface value?
Answer:
\( \frac{R^2}{(R+h)^2} = \frac14 \Rightarrow R+h = 2R \)
h = R (about 6400 km).

Worked example 2. A body weighs 90 N on the surface of the earth. What is its weight at a depth equal to half the earth's radius (uniform density)?
Answer:
\( g_d = g\left(1-\frac{d}{R}\right) = g\left(1-\frac12\right) = \frac g2 \)
Weight \( = 90 \times \frac12 = \) 45 N.