Chapter 14, Problem 43A

To determine

The satellite’s kinetic energy at another point in its orbit.

Answer to Problem 43A

The kinetic energy at another point in its orbit is $3500\text{ MJ}$ .

Explanation of Solution

Given:

The gravitational potential energy at a particular point is $P{E}_{\text{initial}}=5000\text{ MJ}$ .

The kinetic energy at a particular point is $K{E}_{\text{initial}}=4500\text{ MJ}$ .

The satellite’s potential energy at another point is $P{E}_{\text{final}}=6000\text{ MJ}$ .

Formula used:

The expression for total energy is,

  $E=PE+KE$

Here, $PE$ is the potential energy and $KE$ is the kinetic energy.

Calculation:

Consider the total energy remains constant when it changes one point to another point.

The satellite’s kinetic energy at a another point using the relation is,

  $\begin{array}{l}{E}_{\text{final}}=E_{\text{initial}}\\ P{E}_{\text{final}}+K{E}_{\text{final}}=P{E}_{\text{initial}}+K{E}_{\text{initial}}\\ K{E}_{\text{final}}=\left(5000\text{ MJ}\right)+\left(4500\text{ MJ}\right)-\left(6000\text{ MJ}\right)\\ K{E}_{\text{final}}=3500\text{ MJ}\end{array}$

Conclusion:

Thus, the kinetic energy at another point in its orbit is $3500\text{ MJ}$ .