Question

A 1750 kg weather satellite moves in a circular orbit with a gravitational potential energy of 1.69 x 10 to the 10th power J. At its location, free fall acceleration is only 6.4m/s squared. How high above Earth's surface is the satellite?

Answer 1

Given data:

The mass of satellite is m=1750 kg.

The gravitational potential enegy of the statellite is GPE=1.69x10¹⁰ J.

The free fall acceleration is g=6.44 m/s².

The expression for the gravitational potential energy of the satellite is given by,

$\text{GPE}=\text{mgh}$

Here, g is the free fall acceleration, and h is the height of satellite above the Earth surface.

Substitute the given values in above equation,

$\begin{gathered} 1.69\times10^{10}=(1750)(6.4)h \\ h=1.49\times10^6\text{ m} \end{gathered}$

Thus, the height of satellite above the Earth surface is 1.49x10⁶ m.