Question

A satellite 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.)

Answer 1

The height from the surface of the earth is  2650 km              

Explanation:

Step 1:

Assume the height is h. (on the surface of the earth)  

We know the formula,  

At the station on earth,

$\begin{equation}F=\frac{G M m}{R^{2}}$

M's the earth's Mass.  

m is the satellite mass

R is the earth's radius.  

Step 2:

So now, because of the earth at which the height exerts on the satellite

$\begin{equation}\begin{aligned}&h=\frac{F}{2}\\&F_{h}=\frac{G M m}{(R+h)^{2}}=\frac{F}{2}\end{aligned}$

or

$\begin{equation}\begin{aligned}&F_{h}=\frac{G M m}{(R+h)^{2}}=\frac{G M m}{2 R^{2}}\\&(R+h)^{2}=2 \mathrm{R}^{2}\\&h^{2}+2 R h+R^{2}=2 R^{2}\\&\mathrm{h}^{2}+2 \mathrm{Rh}+\mathrm{R}^{2}-2 \mathrm{R}^{2}=0\\&\mathrm{h}^{2}+2 \mathrm{Rh}-\mathrm{R}^{2}=0\\&\mathrm{h}=\frac{(-2 \mathrm{R}+2 \mathrm{R} \sqrt{2})}{2}\\&\mathrm{h}=\frac{2 \mathrm{R}(-1+\sqrt{2})}{2}\\&\mathrm{h}=\mathrm{R}(\sqrt{2}-1)\end{aligned}$

R=6400

$\begin{equation}\begin{aligned}&\mathrm{h}=6400(\sqrt{2}-1)\\&\mathrm{h}=6400(1.414-1)\\&\mathrm{h}=6400(0.414)\\&\mathrm{h}=2649.6 \mathrm{km}\end{aligned}$

h = 2649.6 km

   = 2650 km  

The height from the surface of the earth is therefore h = 2650 km              

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

2650 Km

will be your answer

hope it help