What's ratio of the orbital period for a satellite in
a geosynchronous orbit around the Earth to that of a satellite in
an orbit slightly above the surface of the Earth? The Earth has a
radius of 6371km, and a satellite in a circular geosynchronous
orbit has a radius of 42000km from the center of the earth. The
mass of the Earth is 5.792*10^24kg, and Newton's gravitation
constant is G=6.674*10^-11N*m^2/kg^2
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...
A geosynchronous orbit is one in which the satellite orbits
above the equator and has an orbital period of 24 hours so that it
is always above the same point on the spinning earth. Calculate the
altitude of such a satellite.
(b) What is the gravitational field experienced by the
satellite? Give your answer as a percentage in relation to the
gravitational eld at the earth's surface.
Instruction:
First, solve the problem in terms of variables. Then,
calculate the numerical...
A geosynchronous communications satellite in stationary orbit is
used to relay radio messages from one point to another around the
earth’s curved surface. Apply Newton’s second law to the satellite
to find its altitude in meters above the earth’s surface.
RE = 6.38 × 106 m,
mE = 5.97 × 1024 kg, and G
= 6.674 × 10−11 Nm2/kg2.
A satellite is launched into an orbit at an altitude 200 km
above the surface. Onboard is an exquisitely sensitive atomic clock
that is synchronized with an identical clock on Earth. After
orbiting for one year, the satellite is captured, returned to
Earth, and the clocks compared. What will be the shift in time
between the two clocks?
A 425 kg satellite is in a circular orbit at an altitude of 400
km above the Earth's surface. Because of air friction, the
satellite eventually falls to the Earth's surface, where it hits
the ground with a speed of 2.00 km/s. How much energy was
transformed to internal energy by means of friction?
J
Earth Satellites
Table of Earth Satellites
ISS
Hubble
GPS
Geosynchronous satellite
Moon
h
(miles); height above earth's surface
248
12550
r
(m)
54.5
3.85 x 108
v(m/s)
7.56 x 103
T(s); period
4.32 x 104
8.62 x 104
days/revolution
1.0
Revolutions/day
1.0