Question

A satellite is in circular orbit at a height of 5.0x 10^5 m above the earth’s surface. The radius of the earth is 6.38x10^6m and the mass of earth is 5.98x10^24kg. a) Find the speed of the satellite. b) Find the gravitational field strength at this altitude

Answer 1

a) 7614.11 m/s

b) 8.427 N/kg

Explanation:

Given:

Height (h) = 5.0 x 10^5 m

Radius of the earth (r) = 6.38 x 10^6m

Mass of earth (M) = 5.98 x 10^24kg

Mass of satellite (m)

a)

We can calculate the speed of the satellite by taking into account the following:

`\begin{gathered} F_g=G\cdot(M\cdot m)/(d^2) \\ \\ F_c=(mv^2)/(d) \end{gathered}`

In this case, the gravitational force and the centripetal force are equal, therefore:

`\begin{gathered} G\cdot(M\cdot m)/(d^2)=(mv^(2))/(d) \\ \\ (GM)/(d)=v^2 \\ \\ v=\sqrt{(GM)/(d)} \\ \\ G=6.67\cdot10^(-11)(m^3)/(kg\cdot s^2) \\ \\ d=r+h=6.38\cdot10^6+5.0\cdot10^5=6380000+500000=6880000\text{ m} \\ \\ \text{ We replacing:} \\ \\ v=\sqrt{(6.67\cdot10^(-11)\cdot5.98\cdot10^(24))/(6880000)}=7614.11\text{ m/s} \end{gathered}`

b)

Now, we calculate the gravitational field at that height, like this:

`\begin{gathered} g=(GM)/(d^2) \\ \\ \text{ we replacing } \\ \\ g=(6.67\cdot10^(-11)\cdot5.98\cdot10^(24))/(6880000^2) \\ \\ g=8.427\text{ N/kg} \end{gathered}`