A satellites is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station?(Radius of the earth is $6400$ km.)

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Solution

Suppose the height is $h$ .
At earth station $F = \frac{G M m}{R^{2}}$
$M =$ mass of earth
$m =$ mass of satellite
$R =$ radius of earth
$F = \frac{G M m}{(R + h)^{2}} = \frac{G M m}{2 R^{2}}$
$\Rightarrow 2 R^{2} = (R + h)^{2} \Rightarrow R^{2} - h^{2} - 2 R h = 0$
$\Rightarrow h^{2} + 2 R h - R^{2} = 0$
$h = \frac{(- 2 R \pm \sqrt{4 R^{2} + 4 R^{2}})}{2} = \frac{- 2 R \pm 2 \sqrt{2 R}}{2}$
$= - R \pm \sqrt{2 R} = R (\sqrt{2} - 1)$
$= 6400 \times (0.414)$
$\Rightarrow h = 2650 k m$

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A satellites is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station?(Radius of the earth is 6400km.)

Suppose the height is h.At earth station F=GMmR2M= mass of earthm= mass of satelliteR= radius of earthF=GMm(R+h)2=GMm2R2⇒2R2=(R+h)2⇒R2−h2−2Rh=0⇒h2+2Rh−R2=0h=(−2R±√4R2+4R2)2=−2R±2√2R2=−R±√2R=R(√2−1)=6400×(0.414)⇒h=2650 km

A satellites is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station?(Radius of the earth is $6400$ km.)