10: An earth satellite remains in orbit at a distance of 11600 km from the...

[daniel_walker_27]

I'll show you how to derive the formula for orbital periods, this should help you understand

You should know that an orbiting satellite experiences a gravitational force and a centripetal force through circular motion.

These two forces MUST be equal to keep the planet in orbit. Thus -

(GMm)/r² = mv²/r

The 'm' cancels on either side, and one 'r' from either side cancels, leaving one 'r' on the LHS. Giving -

v² = GM/r
so
v = ?(GM/r)

From circular motion, the velocity = 2 ? f r = (2?r)/T [Where T is the time period]

Substituting gives -

(2?r)/T = ?(GM/r)

If you square both sides you get -

(4?²r²)/T² = GM/r

Rearrange to get in terms of T -

T² = (4?²r³)/(GM) [Just incase it's too small, there is now a cubic power for the 'r']

Therefore -

T = ?(4?²r³) / (GM)

Substitute your values in, making sure to covert the radius in to 'm', and you should get an answer of -

12429.5s = 207.2 minutes (Approx.)

Hope this helps

[KhaliS]

...center of...? An earth satellite remains in orbit at a
distance of 11600 km from the center of the
earth.
What is its period? The universal gravi-
tational constant is 6.67 × 10-11 N · m2/kg2
and the mass of the earth is 5.98 × 1024 kg.
Answer in units of s.