Question

Please help with 1C....

Question 1 (6 pts): Many different types of weather and communications satellites are intentionally placed in “geostationary” orbits, i.e. the satellite will stay above the same place on the earth as the earth rotates.

a. Using Newton’s law of gravity, derive a symbolic expression (no numbers, only variables!) for the altitude of a geostationary orbit in terms of G, M, the mass of the earth, and T, the length of one day, assuming that the orbit is perfectly circular.

b. Now plug in numbers. What is this altitude in meters? What fraction of the way to the moon is this?

c. The moon is “tidally locked” to the earth, which means that the same side of the moon always faces the earth. Another way to say this: the moon rotates around its axis in the same period with which it orbits the earth. Let’s say NASA wanted to park a satellite in circular lunar-stationary orbit above the far side of the moon. Is there a way to do this? Draw a picture of how the moon and this satellite might orbit in relation to the earth. Set up an equation (but do not solve) that would define the altitude of such an orbit.

Answer 1

We have to launch the satellite with very high speed till the force of gravity due to the Earth becomes less, then we can reduce the speed to bind the satellite in the lunar-stationary orbit.

This orbit can be calculated by solving the equation of force

where,

h = height of the satellite from the surface of the moon

v = velocity of the satellite

but,

where,

T = period of the satellite (same as that of the Moon and the Earth)

This is a distance of the satellite from the center of the Moon.

As shown in the diagram, the moon is in between the Earth and the satellite.(as required).

So, we have to add the Earth-Moon distance to (2), to get the total Earth-satellite distance.

i.e.