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The total energy of a satellite is -

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Important Questions on Gravitation

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A satellite is orbiting the Earth in a circular orbit of radius R . Which one of the following statements is true?
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A satellite of mass m is orbiting the earth (of radius R ) at a height h from its surface. The total energy of the satellite in terms of g o , the value of acceleration due to gravity at the earth's surface, is
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If the axis of rotation of the earth were extended into space then it would pass close to -
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A test particle is moving in a circular orbit in the gravitational field produced by a mass density ρr=Kr2. Identify the current relation between the radius R of the particle’s orbit and its period T:
EASY
A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25×106 m above the surface of earth. If earth's radius is 6.38×106 m and g=9.8 s-2, then the orbital speed of the satellite is:
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An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the centre of the earth, it has a speed of 12 km s-1. Neglecting the effect of earth's atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km s-1) ? Give your answer to the nearest integer in km s-1__________.
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The ratio of the binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height h above the earth's surface is (R= radius of the earth).
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A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is: 
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What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R ?
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The minimum number of geostationary satellites required for uninterrupted global coverage is
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A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?
[Given: Mass of planet  =8×1022 kg ,
Radius of planet =2×106 m,
Gravitational constant  G=6.67×10-11 Nm2/kg2 ]
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Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, KAKB is:
EASY
A satellite is orbiting close to the earth and has a kinetic energy K. The minimum extra kinetic energy required by it to just overcome the gravitation pull of the earth is
MEDIUM
A satellite is revolving in a circular orbit at a height h from the earth's surface (radius of earth R; h << R ). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to (Neglect the effect of atmosphere.)
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The mass density of a spherical galaxy varies as Kr over a large distance r from its center. In that region, a small star is in a circular orbit of radius R. Then the period of revolution,T depends on R as:
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In an earth satellite moving in a circular orbit, a piece of metal (weighing 0.016 kg on the earth) is weighed by a spring balance while the metal is suspended in water. If the density of metal is 8 times of water density, then the recorded weight will be (g is acceleration due to gravity)
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A planet is moving in a circular orbit. It completes 2 revolutions in 360 days. What is its angular frequency?
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The relative uncertainty in the period of a satellite orbiting around the earth is 10-2 . If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is:
MEDIUM
The energy required to take a satellite to a height h above the Earth surface (radius of Earth =6.4×103 km ) is E1, and the kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is
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Two stars of masses 3×1031 kg each, and at distance 2×1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is ( Take Gravitational constant G=6.67×10-11 N m2 kg-2 )