Question

where G is a constant and M is the mass of the earth. Calculate the work done by the force of gravity on a particle of mass m as it moves radially from 7500 km to 9400 km from the center of the earth.

Answer 1

Answer:

$-2.0213\times 10^{-7}GMm\text{ J}$

Step-by-step explanation:

Since, the force of gravity is,

$F = -\frac{GMm}{r^2}$

Where,

G = gravitational constant,

M = mass of earth,

m = mass of the particle,

r = distance of particle from centre of the earth,

∵ 7500 km = $7.5\times 10^6$ meters

9400 km = $9.4\times 10^6$ meters

Thus, work done by the force of gravity,

$W=\int_{7.5\times 10^6}^{9.4\times 10^6}F. dr$

$=-\int_{7.5\times 10^6}^{9.4\times 10^6}\frac{GMm}{r^2}dr$

$=GMm[\frac{1}{r}]_{7.5\times 10^6}^{9.4\times 10^6}$

$=GMm(\frac{1}{9.4\times 10^6}-\frac{1}{7.5\times 10^6})$

$=GMm(\frac{7.5-9.4}{9.4\times 10^6})$

$=-GMm(\frac{1.9}{9.4\times 10^6})$

$\approx -2.0213\times 10^{-7}GMm\text{ J}$

Where,

$G = 6.67408\times 10^{-11} \text{ }m^3 kg^{-1} s^{-2}$

$M=5.972\times 10^24 \text{ kg}$