At what height from the surface of earth the total energy of a satellite is equal to its potential energy at a height of 2R from surface of earth ? (R = radius of earth)

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Solution

The correct option is B

Potential energy at a height of 2R from surface of earth

=- $\frac{G M m}{3 R}$

and total energy of a satellite at a height

h=- $\frac{G M m}{2 (R + h)}$

given that - $\frac{G M m}{3 R}$ = - $\frac{G M m}{2 (R + h)}$

$∴$ h= $\frac{R}{2}$

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At what height from the surface of earth the total energy of a satellite is equal to its potential energy at a height of 2R from surface of earth ? (R = radius of earth)

The correct option is B Potential energy at a height of 2R from surface of earth =- GMm3R and total energy of a satellite at a height h=-GMm2(R+h) given that - GMm3R = -GMm2(R+h) ∴ h=R2

The correct option is B

Potential energy at a height of 2R from surface of earth

=- $\frac{G M m}{3 R}$

and total energy of a satellite at a height

h=- $\frac{G M m}{2 (R + h)}$

given that - $\frac{G M m}{3 R}$ = - $\frac{G M m}{2 (R + h)}$

$∴$ h= $\frac{R}{2}$