How to find the distance of a satellite with respect to the moon

[starplaya]

Homework Statement

How far above the surface of the moon should a satellite be placed so that it is stationary with respect tot the moons surface.

Radius of moon = 1737.4 kilometers
mass of moon = 7.36 x 10^22 kilograms
Gravitational Constant = 6.67 x 10^ -11 N x m^2/kg6@

Homework Equations

w = 2pi / T = 2pi/ 2332800 seconds (27 days/period of the moon)
v = w x Radius of satellite
(Ms) Vs^2/Rs = G Mm/Rs^2
Vs^2/Rs = (Gn) Mm/ Rs^2)

Mm = mass of moon
Rm = radius of moon
Ms = mass of satellite
Rs = Radius of satellite
etc...

The Attempt at a Solution

w = 2pi/ 233280 = 2.7 x 10^-6
I'm completely lost after this step. It seems like I don't have enough information

 

[HallsofIvy]

Well, that's all the information you could have isn't it?

The way I would do it (because I'm more mathematics than physics) is set up parametric equations for the satellite's orbit:
$x= r cos(\omega t)$
$y= r sin(\omega t)$

The velocity of the satellite is given by
$v_x= -r\omega sin(\omega t)$
$v_y= r\omega cos(\omega t)$

And the acceleration by
$a_x= -r\omega^2 cos(\omega t)$
$a_y= -r\omega^2 sin(\omega t)$

And now the magnitude of the acceleration is the same as the force of gravity on the satellite due to the moon (divided by the mass of the satellite):
$$\frac{GM}{r^2}$$
where G is the "universal gravitational constant) (NOT "g", the acceleration due to gravity on the Earth's surface) and M is the mass of the moon. The mass of the satellite is, of course, irrelevant.

 

[starplaya]

Well, that's all the information you could have isn't it?

The way I would do it (because I'm more mathematics than physics) is set up parametric equations for the satellite's orbit:
$x= r cos(\omega t)$
$y= r sin(\omega t)$

The velocity of the satellite is given by
$v_x= -r\omega sin(\omega t)$
$v_y= r\omega cos(\omega t)$

And the acceleration by
$a_x= -r\omega^2 cos(\omega t)$
$a_y= -r\omega^2 sin(\omega t)$

And now the magnitude of the acceleration is the same as the force of gravity on the satellite due to the moon (divided by the mass of the satellite):
$$\frac{GM}{r^2}$$
where G is the "universal gravitational constant) (NOT "g", the acceleration due to gravity on the Earth's surface) and M is the mass of the moon. The mass of the satellite is, of course, irrelevant.

I am still having a lot of trouble figuring this particular question out.

 

[phyzguy]

What's the rotation rate of the moon?

 

[starplaya]

27 days

 

[starplaya]

I really have been putting in a ton of effort into this question folks. I spent 30min of the hour and a half it took me to do all of my physics on this one question.

Please, any detailed help would be greatly appreciated

 

[gneill]

Hint: Equate gravitational acceleration to centripetal acceleration.

You'll need to determine an expression for either the angular velocity or linear velocity to use with the centripetal acceleration expression (depending upon whether you use the angular or linear tangential form of the centripetal acceleration formula).

 

[Andrew Mason]

Follow Gneill's advice. Assume that the centre of rotation of the satellite is the centre of mass of the moon (it is really the centre of mass of the moon and satellite, but that is very, very close to the cm of the moon).

Then it is just a matter of plugging in your numbers to determine the radius or orbit.

AM

 

[starplaya]

Don't I at least need the velocity. I mean I can see myself deriving everything with slightly more information, but with absolutely nothing given to me, I'm stunned.

 

[phyzguy]

Where are you stuck? You've written down the correct equations. You know the velocity in terms of w(which you know) and Rs. So the only thing missing is Rs. Substitute vs = w Rs into your last equation and solve for Rs.

 

[NascentOxygen]

Don't I at least need the velocity

Yes. You know the period of rotation, in seconds, and radius is what you're working at finding, so keep it at R_s.

Vs^2/Rs = (Gn) Mm/ Rs^2

I don't know what the n signifies, but that looks like the best of your equations, and you'll recognize that the only actual unknown is Rs.

It's a good idea to carry the units along all the way, with every equation, so you keep a check that you are adhering to the correct units.

 

[yands]

Think as you set on a long stick and ur friend is away from you who is setting on the stick too the stick is fixed in the middle and the ends oscillate up and down like kids game in the garden.
You will see him stationary. Because you both are moving with same angular velocity. So ! Use this logic to solve ur problem

 

[SteamKing]

OP is almost 2 years old. I doubt if the poster has been waiting around for further replies.