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A satellite has a mass of 5750 kg and is in a circular orbit 3.6 * 10^5 m above the surface of a planet. The period of the orbi...Asked by ajo
A satellite has a mass of 5850kg and is in a circular orbit 4.1*10^5m above the surface of a planet. The period of the orbit is two hours. the radius of the planet is 4.15*10^6m.What is the true weight of the satellite when it is at rest on the planet's surface?
can anyone please give me some hints to do it? THANKS A LOT!
can anyone please give me some hints to do it? THANKS A LOT!
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Answered by
drwls
The weight of the satellite at the surface of the planet is
GmM/R^2, where
G is the universal constant of gravity. (Look it up if you don't know it);
m is the satellite's mass (5850 kg)
R is the planet's radius (4.15*10^6 m)
M is the mass of the planet.
You will need to use Kepler's third law, the orbital period and radius to get the planet's mass, M. Remember that the orbit's radius is the sum of 4.1*10^5m (the altitude)and 4.15*10^6 m (the planet's radius)
GmM/R^2, where
G is the universal constant of gravity. (Look it up if you don't know it);
m is the satellite's mass (5850 kg)
R is the planet's radius (4.15*10^6 m)
M is the mass of the planet.
You will need to use Kepler's third law, the orbital period and radius to get the planet's mass, M. Remember that the orbit's radius is the sum of 4.1*10^5m (the altitude)and 4.15*10^6 m (the planet's radius)
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