Question

What is the speed, period, and radial acceleration of a communications satellite in a circular orbit 2000km above the surface of the earth. The earth's radius is 6380 km and its mass is 5.98 x 10^(24)kg. G= 6.67 x10^(-11)Nm^(2)kg^(-2). At what height above the surface of te earth should the satellite be placed for the orbital speed to be 4,000m/s^(2)?

Please just solve this for me and explain step by step. Thank you!

Answer 1

if you look at the summation of forces on the satellite, you can see that the only force acting on the satellite is the force of gravity - and since it's in a circular orbit, you know that the satellite is undergoing centripetal acceleration, so

\[\sum F = F_g=ma_{cent}\]
Looking at the equation for the force of gravity on the satellite is
\[| \textbf F| = G\frac{mM_E}{R_S^2}\]
where RS is the distance from the center of the earth to the satellite, ME is the mass of the earth, and m is the mass of the satellite.
Now looking at the centripetal acceleration: by definition, it's
\[a_{cent} = \frac{v^2}{R_S}\]
where v is the velocity of the satellite, and RS is, again, the distance from the center of the earth to the satellite.

Equating gives
\[G\frac{mM_E}{R_S^2}=m\frac{v^2}{R_S}\]
The m's cancel out, and rearranging leaves you with
\[R_S = G\frac{M_E}{v^2}\]

Then to find the distance from the surface of the earth

\[R=R_S-R_E\]

Answer 2

Wow, you did an amazing job on this @AllTehMaffs! I gave you a medal too (and so did a lot of other people)!

Answer 3

Yes. Especially well done.