CBSE
Calculate the distance below and above the surface of the earth, at which the value of acceleration due to gravity becomes 1/4th that at earth's surface ?
The height vertically above the earth's surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the earth)
8 R
9 R
10 R
20 R
The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the Earth is (Here R is the radius of the earth)
nmgR
If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass m raised from the earth's surface to a height equal to the radius R of the earth is
mgR
2mgR
Average distance of the Earth from the Sun is L1. If one year of the Earth = D days, one year of another planet whose average distance from the Sun is L2 will be
A stone falls freely under gravity. It covers distances h1, h2, and h3, in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between h1, h2, and h3 is
h1 = 2h2 = 3h3
h2 = 3h1 and h3 = 3h2
h1 = h2 = h3
A body of mass m taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be
mg2R
3 mgR
Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1 m, 2 m, 4 m, 8 m respectively from the origin. The resulting gravitational potential due to this system at the origin will be
− G
− 4 G
An artificial satellite moves in a circular orbit around the earth. Total energy of the satellite is given by E. The potential energy of the satellite is
- 2E
2E
Two particles of mass m1 and m2 approach each other due to their mutual gravitational attraction only. Then
accelerations of both the particles are equal
acceleration of the particle of mass m1 is proportional to m1
acceleration of the particle of mass m1 is proportional to m2
acceleration of the particle of mass m1 is inversely proportional to m1