Question

4. A 1,750 kg weather satellite moves in a circular orbit with a gravitational

Answer 1

Explanation:

It is known that the value of free fall acceleration is g = 6.44 $m/s^{2}$. And, the value of radius of Earth is $6.4 \times 10^{6} m$.

Let us assume that height of the satellite is h.

It is known that,

             g = $\frac{GM}{r^{2}}$        

where,   r = (R + h)

Hence,  

              r = $\sqrt{\frac{GM}{g}}$

              r = $\sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{6.44 m/s}}$

                = $7.9 \times 10^{6} m$

Now, formula for height of satellite above the Earth's surface is as follows.

            h = r - R

               = $7.9 \times 10^{6} - 6.4 \times 10^{6}$

               = $1.5 \times 10^{6} m$

Thus, we can conclude that the satellite is $1.5 \times 10^{6} m$ high above Earth's surface.