How Does Satellite Orbit Duration Affect Gravity Calculation?

[Physics103]

A satellite orbits Earth with T= 5.1 x 10^3 s. What is the value for g at this point.

some constants i found:
Mass of Earth = 5.98 x 10^24lg
r of Earth = 6.38 x10^6m

I realize that because the T is pretty small that the value for g will be close to 9.8. This is what i can think of

Gmemo / r^2 = mg
Gme / r^2 = g

and

Fc = mo4pie^2r / T^2
Since Fc = Fg

mg = mo4pie^2r / T^2
thus g = 4pie^2r / T^2

i tried subbing the equations together to get r and from that get g but it seems to me like I am doing it wrong. Its only a multiple choice problem so i don't see why it would be so complicated (if infact its done they i showed).

Thank for any help

 

[Smilecdl]

The gravitational constant (g) depends on the square of the distance from the centre of the Earth.

 

[Physics103]

yes but your not given the distance... only the radius of earth

 

[vaishakh]

Let the force of attraction between the satelite be F. Let the mass of Earth be taken as M and G is constant of gravity while g is thevalue of acceleration due to gravity for a point situated there at a point r distance away from the surface of the Earth which has radius as R. So distance between centre of mass is R + r. Let the mass of the staelite be m.
So now by Newton's law we get, F = mg = mMG/(R + r)^2
Therefore g = MG/(R + r)^2. In this equation we have two unknowns and so the question is unsolvable from just this equation.
So let us take the equation involving the period taken to revolve around earth. It took 5100s to complete one revolution. Fromthis information, by assuming that the only Force applied on the on object is the Earth's gravitational speed while other forced are negligibe compared to this, we can say that there is no tangential acceleration along the circular path propelled by the body. Thus we get the angular velocity of the body. And from this we get the radial component of acceleration of the circle.
w = 2pi/T
F = mw^2/R + r.
g = w^2/R + r
MG/(R + r)^2 = w^2/(R + r)
MG/w^2 = R + r
Even if it is a multiple choice question I cannot think of something that will help you in this question even without solving it.

 

[Physics103]

ahh my class skipped angular velocity maybe that's why i didn't know what to do.

thanks a bunch

 

[andrevdh]

Using the centripetal force equation one gets
$$F_c=\frac{mv^2}{r}$$
and substituting for v
$$v=\frac{2\pi r}{T}$$
this centripetal force is supplied by the universal gravitational force and by using the fact that
$$g=G\frac{M}{r^2}$$
you get the required formula in terms of $r,\ T$