The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R
from the surface of the earth is (g = acceleration due to gravity at the surface of the
earth)
1.
2.
3.
4. g
If V, R, and g denote respectively the escape velocity from the surface of the earth, the
radius of the earth, and acceleration due to gravity, then the correct equation is:
1.
2. V=
3. V = R
4. V=
The depth at which the effective value of acceleration due to gravity is , is:
1. R
2.
3.
4.
The value of \(g\) at a particular point is \(9.8~\text{m/s}^2\). Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass then value of \(g\) at the same point will now become: (assuming that the distance of the point from the centre of the earth does not shrink)
1. | \(4.9~\text{m/s}^2\) | 2. | \(3.1~\text{m/s}^2\) |
3. | \(9.8~\text{m/s}^2\) | 4. | \(19.6~\text{m/s}^2\) |
1. | decrease by \(1\%\) | 2. | increase by \(1\%\) |
3. | increase by \(2\%\) | 4. | remain unchanged |
The earth (mass = ) revolves round the sun with angular velocity
in a circular orbit of radius . The force exerted by the sun on
the earth in Newtons, is
1.
2. Zero
3.
4.
The gravitational force between two point masses and at separation r is given by The constant k
1. Depends on system of units only
2. Depends on medium between masses only
3. Depends on both 1 and 2
4. Is independent of both 1 and 2
The density of a newly discovered planet is twice that of earth. The acceleration due to
gravity at the surface of the planet is equal to that at the surface of the earth. If the
radius of the earth is R, the radius of the planet would be
1. 2R
2. 4R
3.
4.
If R is the radius of the earth and g the acceleration due to gravity on the earth's surface,
the mean density of the earth is:
1.
2.
3.
4.
A planet has twice the radius but the mean density is as compared to earth. What is the ratio of escape velocity from earth to that from the planet ?
1. 3:1
2. 1:2
3. 1:1
4. 2:1