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qUESTIONS

The height above the earth’s surface at which the acceleration due to gravity becomes (1/n) times the value at the surface is (R = radius of the earth)

Earth revolves round the sun in a circular orbit of radius ‘R’. The angular momentum of the moving earth is directly proportional to:

A launching vehicle carrying an artificial satellite of mass ‘m’ is set for launch, on the earth’s surface. If the satellite is intended to move in a circular orbit of radius ‘7R’, the minimum energy required to be spent by the launching vehicle is: (G= Universal gravitation constant, M and R be the mass and radius of the earth)

If the radius of a planet is ‘R’ and density is ‘p’, the escape velocity ‘v’ of any body from its surface will be:

A satellite of mass ‘m’ revolving around the earth of radius ‘r’ has kinetic energy ‘E’. Its angular momentum is:

A body weight ‘W’ newton on the surface of the earth. Its weight at a height equal to half the radius of the earth, will be:

If ‘M’ and ‘R’ are mass and radius of earth respectively and ‘m’ is mass of a satellite revolving in the circular orbit of radius ‘r’ at height ‘h’ from the surface of earth. (G= Gravitational constant) Match column I with column II.

The acceleration due to gravity on moon is 1/6 times the acceleration due to gravity on earth. If the ratio of the density p to the density of moon p is 5/3, then the radius of moon R in terms of the radius of earth R is:

A spring balance is graduated on sea level. If a body is weighed with this balance at consecutively increasing heights from earth’s surface, the weight indicated by the balance

A research satellite of mass 200 kg circles the earth in an orbit of average radius 3R/2, where R is the radius of the earth. Assuming the gravitational pull on a mass of 1 kg on the earth’s surface to be 10 N, the pull on the satellite will be

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