In: Physics
A geosynchronous satellite is one that stays above the same point on the equator of the earth. Determine the height above the Earth's surface such a satellite must orbit and find it's speed. (Note that the radius in equation is measured from the center of the earth, not the surface.) You may use the following constants:
The universal gravitational constant G is 6.67 x 10-11 N m2 / kg2 .
The mass of the earth is 5.98 x 1024 kg.
The radius of the earth is 6.38 x 106 m.
Hint: In order for the satellite to remain in orbit, the gravitational force must equal the centripetal force:
universal gravitational constant G is 6.67 x 10-11 N m2 / kg2
The mass of the earth is 5.98 x 1024 kg
The radius of the earth is 6.38 x 106 m.
We assume that the satellite is in Uniform Circular
Motion (UCM). If this is true, the acceleration will be given
by
a = v^2/r. (1)
This is the "a" in the force equation,
F = ma. (2)
However, the force on the satellite is also given by the
gravitational force it due to the Earth:
F = GMm/(r^2) (3)
Combining (2) and (3) above, we get
Gm/r = v^2 (4)
Since the satellite goes around the Earth once each day (T = 1 day
= 86400 s), its speed is
v = 2*pi*r/T (5).
Plugging (4) into (5) gives us
GM/r = (2*pi*r)^2/(T^2) (6).
We want to solve this for r, so we have
r^3 = GMT^2/[4*(pi)^2]. (7)
It's plug-and-chug time. Substituting values for G, M = 5.98 x
10^24 kg, and T, we get
= 7.54*10^22
r = 4.23 x 10^7 m. (8).
The Earth's radius is 0.64 x 10^7 meters, so that the answer is
3.59 x 10^7 m for the height above the surface.
Using the value from (8) in equation (4) and solving for v, we find
that
v = 3070 m/s (9)