Question

Question 11 of 30A satellite with a mass of 100 kg fires its engines to increase velocity, therebyincreasing the size of its orbit about Earth. As a result, it moves from acircular orbit of radius 7.5 x 106 m to an orbit of radius 7.7 x 106 m. What isthe approximate change in gravitational force from Earth as a result of thischange in the satellite's orbit? (Recall that Earth has a mass of 5.97 x 1024 kgand G = 6.67 x 10- 11 N·m2/kg)O A. -8NB. -36 NC. -24 ND. -16 NSUBMIT

Answer 1

Given:

The mass of the satellite is m = 100 kg

The radius of the lower orbit is

$R_l=\text{ 7.5}\times10^6\text{ m}$

The radius of the higher orbit is

$R_h=7.7\times10^6\text{ m}$

The mass of the earth is

$M\text{ = 5.97}\times10^{24}\text{ kg}$

The universal gravitational constant is

$G\text{ = 6.67}\times10^{-11}\text{ Nm}^2\text{ /kg}^2$

To find the approximate change in gravitational force.

Explanation:

The gravitational force due to the lower orbit is

$\begin{gathered} F_l=\frac{GMm}{(R_l)^2} \\ =\frac{6.67\times10^{-11}\times5.97\times10^{24}\times100}{(7.5\times10^6)^2} \\ =\text{ 707.9 N} \end{gathered}$$\begin{gathered} F_l=\frac{GMm}{(R_l)^2} \\ =\frac{6.67\times10^{-11}\times5.97\times10^{24}\times100}{(7.5\times10^6)^2} \\ =\text{ 707.9 N} \end{gathered}$$\begin{gathered} F_l=\frac{GMm}{(R_l)^2} \\ =\frac{6.67\times10^{-11}\times5.97\times10^{24}\times100}{(7.5\times10^6)^2} \\ \approx\text{ 707.9 N} \end{gathered}$

The gravitational force due to the higher orbit is

$\begin{gathered} F_h=\frac{GMm}{(R_h)^2} \\ \approx671.61\text{ N} \end{gathered}$

The approximate change in gravitational force is

$\begin{gathered} \Delta F=F_h-F_l \\ =671.61-707.9 \\ =-36.29\text{ N} \end{gathered}$

The approximate change in gravitational force is 36.29 N