what is the mass of the planet that has a satellite whose time period is T and orbital radius is r?
Answers
Answer:
Explanation:
It was Copernicus who, first of all, introduced the idea that the central body of our planetary system was Sun rather than Earth. Kepler later on confirmed by putting it on a solid mathematical back ground. Kepler announced two of his laws in 1609 and the third one 10 years later. The laws can be stated as below:
(a) Kepler’s first law (law of elliptical orbit):- A planet moves round the sun in an elliptical orbit with sun situated at one of its foci.
(b) Kepler’s second law (law of areal velocities):- A planet moves round the sun in such a way that its areal velocity is constant.
(c) Kepler’s third law (law of time period):- A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit.
T2 ∝ r 3
Here R is the radius of orbit.
T2 = (4π2/GM) r 3
Earth and Its Satellite
Consider a satellite of mass m revolving in a circular orbit around the Earth, which is located at the centre of its orbit. If the satellite is at a height h above the Earth's surface, the radius of its orbit r = Re + h, where Re is the radius of the Earth. The gravitational force between Me & m provides the centripetal force necessary for circular motion, i.e.,
GMem/(Re+h)2 = mv2/(Re+h)
Or v2 = GMe/(Re+h) or v = √GMe/(Re+h)
Hence orbital velocity depends on the height of the satellite above Earth's surface. Time period T of the satellite is the time taken to complete one revolution.
Therefore, T = 2Πr/v = 2Π(Re + h)√(Re+h)/GMe
or T2 = 4Π2(Re+h)3/GMe where r = Re + h
If time period of a satellite is 24 hrs. Then,
r = [GMeT2/4Π2]1/3 = 42400 km and h = 36000 km.
This gives the height of a satellite above the Earth's surface whose time period is same as that of Earth's. Such a satellite appears to be stationary when observed from the Earth's surface and is hence known as Geostationary satellite.
For a satellite very close to the surface of Earth i.e. h << Re then
r ≈ Re
vorbital = √GMe/Re = √gRe
Orbital Velocity
Simulation for Kepler Laws of Motion
Satellite in circular orbit:-
For different velocities, the trajectory of the satellite would be different. Let us consider these cases.
If v is the velocity given to a satellite and vO represents the velocity of a circular orbit and ve the escape velocity.
i.e. v0 = √GMe/(Re+h)