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

`-2.0213* 10^(-7)GMm\text{ J}`

Explanation:

Since, the force of gravity is,

`F = -(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* 10^6` meters

9400 km =
`9.4* 10^6` meters

Thus, work done by the force of gravity,

`W=\int_(7.5* 10^6)^(9.4* 10^6)F. dr`

`=-\int_(7.5* 10^6)^(9.4* 10^6)(GMm)/(r^2)dr`

`=GMm[(1)/(r)]_(7.5* 10^6)^(9.4* 10^6)`

`=GMm((1)/(9.4* 10^6)-(1)/(7.5* 10^6))`

`=GMm((7.5-9.4)/(9.4* 10^6))`

`=-GMm((1.9)/(9.4* 10^6))`

`\approx -2.0213* 10^(-7)GMm\text{ J}`

Where,

`G = 6.67408* 10^(-11) \text{ }m^3 kg^(-1) s^(-2)`

`M=5.972* 10^24 \text{ kg}`