The gravitational force of the Earth on a roc | vedclass

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(A) The gravitational force at the surface of the Earth is $F_s = \frac{GM_e m}{R_e^2}$.The gravitational force at height $h$ is $F_h = \frac{GM_e m}{

Vedclass · Accepted Answer

$h = R_e(\sqrt{3} - 1)$

Vedclass · Answer

(A) The gravitational force at the surface of the Earth is $F_s = \frac{GM_e m}{R_e^2}$.The gravitational force at height $h$ is $F_h = \frac{GM_e m}{(R_e + h)^2}$.According to the problem,$F_h = \frac{1}{3} F_s$.Substituting the expressions: $\frac{GM_e m}{(R_e + h)^2} = \frac{1}{3} \frac{GM_e m}{R_e^2}$.Simplifying the equation: $(R_e + h)^2 = 3 R_e^2$.Taking the square root on both sides: $R_e + h = \sqrt{3} R_e$.Therefore,$h = \sqrt{3} R_e - R_e = R_e(\sqrt{3} - 1)$.

The gravitational force of the Earth on a rocket at a height $h$ from the surface of the Earth is $\frac{1}{3}$ of the gravitational force on the surface of the Earth. Find the relationship between $h$ and $R_e$ (the radius of the Earth).

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