the gravitational force acting on a rocket at a height h from the earth surface is 1/3 of the force acting on a body at sea level What is relation b/w h and R (Radius of the earth)

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Solution

Gravitational force exerted on rocket or man m is given by
$F = G \frac{m M_{e}}{r^{2}}$
r is the distance of the rocket from the center of earth at sea level
$r = R$
$3 G \frac{m M_{e}}{(R + H)^{2}} = G \frac{m M e}{R^{2}}$
$3 R^{2} = (R + H)^{2} \Rightarrow R + H = \sqrt{3 R}$
$∴ H = (\sqrt{3} - 1) R = 0.73 R$

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the gravitational force acting on a rocket at a height h from the earth surface is 1/3 of the force acting on a body at sea level What is relation b/w h and R (Radius of the earth)

Gravitational force exerted on rocket or man m is given byF=GmMer2r is the distance of the rocket from the center of earth at sea levelr=R3GmMe(R+H)2=GmMeR23R2=(R+H)2⇒R+H=√3R∴H=(√3−1)R=0.73R