Question

Given: G = 6.67259 × 10−11 Nm2/kg2
The acceleration of gravity on the surface of
a planet of radius R = 7770 km is 12.6 m/s2.
What is the period T of a satellite in circu-
lar orbit h = 10800.3 km above the surface?
Answer in units of s. *note:*(must convert distance to meters)

Answer 1

\[mg = G\frac{ mM }{R^2 }\]

Answer 2

\[g = G\frac{ M }{R^2 }\]
\[12.6=6.67259 × 10^{−11}\frac{ M }{(7700*1000)^2 }\]

Answer 3

I dont have mass thats the part that is tripping me up lol

Answer 4

find M, the mass of the planet, and use in:
\[\frac{ V^2 }{ r} = G\frac{ M }{ r^2 }\]
(where r is the radius of the orbit) in order to find V.
from V, period is easy to find...

Answer 5

V=\[\sqrt{gr}\]

Answer 6

then i took that and used T=2piR/V

Answer 7

then T/3600

Answer 8

should i have divided by something else to get my period? I:

Answer 9

doesn't work that way

Answer 10

the value given for 'g' is only valid on the planet's surface

Answer 11

it's a crazy fact about gravity that the force of attraction two masses experience is dependent on the inverse square of the distance between them.... if you were on the moon, do you think you would experience an acceleration of 9.8 m/s^2 towards the earth?

Answer 12

this is universal gravitation

Answer 13

so g is constant for all space, is that what you're saying? ... cool.

Answer 14

can i try to solve it

Answer 15

actually my equation editor is not working now. its a big problem

Answer 16

note that \(g=\frac{GM}{r^2}\)
g=12.6, G is given, R is 7770000m.
I think you can find the M......