Physics > Gravitation > 2.0 Variation of acceleration due to gravity

  Gravitation
    1.0 Newton's law of gravitation
    2.0 Variation of acceleration due to gravity
    3.0 Gravitational field
    4.0 Gravitational potential
    5.0 Gravitational potential energy
    6.0 Satellites
    7.0 Kepler's law of planetary motion
    8.0 Problem solving technique

2.3 Variation of acceleration due to gravity $(g)$ due to depth


Let the particle of mass $m$ is kept at a depth $d$ from the surface of earth. So, the gravitation force exerted on the particle of mass $m$ is,
$$F = \frac{{GM'm}}{{{{\left( {R - d} \right)}^2}}} \quad ...(i)$$
Density of earth is, $$\rho = \frac{M}{{\left( {\frac{4}{3}\pi {R^3}} \right)}}$$
Then mass $M$ is given by, $$M' = \rho V$$$$M' = \frac{M}{{\left( {\frac{4}{3}\pi {R^3}} \right)}} \times \frac{4}{3}\pi {\left( {R - d} \right)^3}$$$$M' = \frac{{M{{\left( {R - d} \right)}^3}}}{{{R^3}}} \quad ...(ii)$$
From equation $(i)$ and $(ii)$ we get,
$$F = \frac{{Gm}}{{{{\left( {R - d} \right)}^2}}} \times \frac{{M{{\left( {R - d} \right)}^3}}}{{{R^3}}}$$$$F = \frac{{GMm\left( {R - d} \right)}}{{{R^3}}}$$$$\frac{F}{m} = \left( {\frac{{GM}}{{{R^2}}}} \right)\left( {\frac{{R - d}}{R}} \right)$$$$g' = g\left[ {1 - \frac{d}{R}} \right]$$
At the centre $(d=R)$ then $g'=0$.

The value of acceleration due to gravity decreases with depth. It is maximum at the earth's surface and minimum at the centre of the earth.

Note:





$$F = \frac{{GMm}}{{{{\left( {R + h} \right)}^2}}}$$$$F = \frac{{GMm}}{{{R^3}}}\left( {R - d} \right)$$
$$g' = g\left[ {1 - \frac{{2h}}{R}} \right]$$$$g' = g\left[ {1 - \frac{d}{R}} \right]$$
If a body is taken above the surface of earth, then the value of acceleration due to gravity varies inversely as the square of the distance from the centre of the earthIf the body is taken inside the earth, acceleration due to gravity decreases linearly with distance from the centre of the earth








$$F = \frac{{GMm}}{{{{\left( {R + h} \right)}^2}}}$$
$$F = \frac{{GMm}}{{{R^3}}}\left( {R - d} \right)$$
$$g' = g\left[ {1 - \frac{{2h}}{R}} \right]$$
$$g' = g\left[ {1 - \frac{d}{R}} \right]$$
If a body is taken above the surface of earth, then the value of acceleration due to gravity varies inversely as the square of the distance from the centre of the earth
If the body is taken inside the earth, acceleration due to gravity decreases linearly with distance from the centre of the earth
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