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}*10^6\text{ m}`

The radius of the higher orbit is

`R_h=7.7*10^6\text{ m}`

The mass of the earth is

`M\text{ = 5.97}*10^(24)\text{ kg}`

The universal gravitational constant is

`G\text{ = 6.67}*10^(-11)\text{ Nm}^2\text{ /kg}^2`

To find the approximate change in gravitational force.

Step-by-step explanation:

The gravitational force due to the lower orbit is

`\begin{gathered} F_l=(GMm)/((R_l)^2) \\ =(6.67*10^(-11)*5.97*10^(24)*100)/((7.5*10^6)^2) \\ =\text{ 707.9 N} \end{gathered}`
`\begin{gathered} F_l=(GMm)/((R_l)^2) \\ =(6.67*10^(-11)*5.97*10^(24)*100)/((7.5*10^6)^2) \\ =\text{ 707.9 N} \end{gathered}`
`\begin{gathered} F_l=(GMm)/((R_l)^2) \\ =(6.67*10^(-11)*5.97*10^(24)*100)/((7.5*10^6)^2) \\ \approx\text{ 707.9 N} \end{gathered}`

The gravitational force due to the higher orbit is

`\begin{gathered} F_h=(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