Having trouble working out centripetal force?

[Jan Berkhout]

Homework Statement

The Russian Mir space station had a mass of 130 tonnes and orbited Earth at an altitude of 480km with an orbital speed of 7621.4m/s. The diameter of Earth is 12 760 km.
a) What centripetal force was acting on it?
b) Find the value of the acceleration due to gravity acting on an astronaut in Mir.

Homework Equations

F_c=mv²/r
a=F/m

The Attempt at a Solution

m = 130000kg r = 6860000m v = 7621.4m/s

So F_c=(130000×7621.4²)/6860000
∴F_c=1100750.136 N

The answers say 30N. What on Earth have I done wrong?

I also do not know how to go about doing b.

 

[Charles Link]

I would recommend using scientific notation, and also only keeping two or three significant figures, but I believe your answer for the force is correct. $\\$ For part (b), you have the correct equation, $a=F/m$. That is the acceleration of the spaceship, but that is the same as the astronaut's acceleration. Because acceleration due to gravity falls off as an inverse square as measured from the center of the earth, since this satellite is in somewhat of a low Earth orbit, your answer should not be tremendously less than $a=g=9.8$ m/sec^2 that you get at the surface of the earth. Try computing it and see what you get. $\\$ Additional note: For the mass in tonnes, I think you are assuming a metric ton.

 

[lekh2003]

The answers say 30N.

This answer seems miraculously wrong.

 

[Jan Berkhout]

I would recommend using scientific notation, and also only keeping two or three significant figures, but I believe your answer for the force is correct. $\\$ For part (b), you have the correct equation, $a=F/m$. That is the acceleration of the spaceship, but that is the same as the astronaut's acceleration. Because acceleration due to gravity falls off as an inverse square as measured from the center of the earth, since this satellite is in somewhat of a low Earth orbit, your answer should not be tremendously less than $a=g=9.8$ m/sec^2 that you get at the surface of the earth. Try computing it and see what you get. $\\$ Additional note: For the mass in tonnes, I think you are assuming a metric ton.

I calulated part b and got 8.47m/s which seems right! I think the sheet has the answers for this question very wrong haha.

This answer seems miraculously wrong.

Yes. It does indeed. I'm just going to go with saying the sheet is insanely incorrect.

 

[Charles Link]

For part (b) make sure you put it in the right units which is 8.47 m/sec^2.