Question

a satellite with a mass of 150 kg fires its engines to increase velocity, thereby increasing the size of its orbit about earth. as a result, it moves from a circular orbit of radius 7.5×10^6 m to an orbit of radius 7.7×10^6 m. what is the approximate change in gravitational force from earth as a result of this change in the satellites orbit? (recall that earth has a mass of 5.97×10^24 kg and g = 6.67×10^(-11) n·m^2/kg^2.)

a. -113 n
b. -148 n
c. -216 n
d. -54 n

Explanation:

Step1: Recall gravitational - force formula

The gravitational - force formula is $F = G\frac{Mm}{r^{2}}$, where $G = 6.67\times10^{-11}\ N\cdot m^{2}/kg^{2}$, $M = 5.97\times10^{24}\ kg$ (mass of Earth), $m = 150\ kg$ (mass of satellite), and $r$ is the radius of the orbit.

Step2: Calculate initial gravitational force $F_1$

For $r_1=7.5\times 10^{6}\ m$, $F_1 = G\frac{Mm}{r_1^{2}}$.
Substitute the values:
\[

$$\begin{align*} F_1&=(6.67\times 10^{-11})\frac{(5.97\times 10^{24})\times150}{(7.5\times 10^{6})^{2}}\\ &=(6.67\times 10^{-11})\frac{8.955\times 10^{26}}{5.625\times 10^{13}}\\ &=\frac{6.67\times8.955\times 10^{-11 + 26}}{5.625\times 10^{13}}\\ &=\frac{59.73985\times 10^{15}}{5.625\times 10^{13}}\\ &=\frac{5973.985}{5.625}\\ &\approx1062.04\ N \end{align*}$$

\]

Step3: Calculate final gravitational force $F_2$

For $r_2 = 7.7\times 10^{6}\ m$, $F_2=G\frac{Mm}{r_2^{2}}$.
Substitute the values:
\[

$$\begin{align*} F_2&=(6.67\times 10^{-11})\frac{(5.97\times 10^{24})\times150}{(7.7\times 10^{6})^{2}}\\ &=(6.67\times 10^{-11})\frac{8.955\times 10^{26}}{5.929\times 10^{13}}\\ &=\frac{6.67\times8.955\times 10^{-11 + 26}}{5.929\times 10^{13}}\\ &=\frac{59.73985\times 10^{15}}{5.929\times 10^{13}}\\ &=\frac{5973.985}{5.929}\\ &\approx1008.77\ N \end{align*}$$

\]

Step4: Calculate the change in gravitational force $\Delta F$

$\Delta F=F_2 - F_1$.
$\Delta F=1008.77 - 1062.04=- 53.27\ N\approx - 54\ N$

Answer:

D. -54 N