Centripetal accelearation?

Question

6.) A 1 500 kg satellite orbits the earth at 2 500 m/s. What is the satellite's centripetal acceleration?

answer: 0.098 m/s^2

how to do this question??

Thanks

Answer 1

If the distance from the earth isn't given, then you need two equations to solve this question.  You need an orbital mechanics equation to solve for the radius of the orbit, and then the equation for centripetal acceleration.

I'm assuming that the orbit of the satellite is circular.  So for the orbital mechanics equation, you have the following:

V = sqrt[(g*(R^2))/r], where g is the acceleration of gravity (9.8 m/s^2), R is the radius of the earth (approx. 6370 km), and r is the distance from the center of the earth.

You need r, so rearranging the equation, we have the following:

r = (g*(R^2))/(V^2) = 63624579.2 m (distance from center of earth = 6370000 m + distance from surface of earth)

Once this is found, you can find the centripetal acceleration:

a = (V^2)/r = 0.098 m/s^2

The MASS of the satellite is NOT needed.  It WOULD be needed if you needed the centripetal force, which would be the mass of the satellite multiplied by the centripetal acceleration.  REMEMBER TO CHECK UNITS!!!

Answer 2

It is helpless to solve this question without the required information that how much distance apart are they i.e. 'r' ????

Then use a = v^2 / r.

Answer 3

My physics knowledge is a tad rusty, but I believe you need to know the radius of the satellite's orbit in order to solve for the centripetal acceleration.  If you need help, try this web page:

http://theory.uwinnipeg.ca/physics/circ/...

Anyhow, by rearranging the following equation:

a[centripetal] = m(((v[tangential])^2) / radius) = 0.098 [m/s^2]

I obtained this as the radius of the orbit:

(9.566 x 10^10) [meters]

I'm not entirely sure that I did it correctly, but that's my take on it.