Chapter 7, Problem 78A

To determine

To compare: Acceleration due to gravity at one Earth’s radius from surface in terms of $g$ .

Answer to Problem 78A

Acceleration due to gravity at $d=2r_E$ is $\frac{1}{4}$ of the value at Earth’s surface.

Explanation of Solution

Given:

Satellite is one Earth’s radius above Earth’s surface.

Formul used:

Now, acceleration at a distance $d$ from Earth’s surface is given as

  $a=g{\left(\frac{r_E}{d}\right)}^2$

Calculation:

Distance of satellite from Earth’s center

  $\begin{array}{c}d=r_E+r_E\\ =2r_E\end{array}$

Here $r_E$ is radius of Earth.

Now, acceleration at a distance $d$ from Earth’s surface is,

  $\begin{array}{c}a=g{\left(\frac{r_E}{d}\right)}^2\\ =g{\left(\frac{r_E}{2r_E}\right)}^2\\ =\frac{g}{4}\end{array}$

Conclusion:

Therefore, acceleration due to gravity at $d=2r_E$ is $\frac{1}{4}$ of the value at Earth’s surface.