A black hole is a body from whose surface nothing may ever escape. What is the condition for a uniform spherical mass M to be a black hole? What should be the radius of such a black hole if its mass is the same as that of the Earth?

The escape velocity from the earth is V. What is the escape velocity from a planet whose mass and radius is twice those of the earth ?

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 xx 10^24 kg, radius = 6400 km

What is gravitational potential energy at a point? How much of work is done in shifting mass from the surface to a height equal to its radius?

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 x 10^24 kg and radius of the earth 6.4 x 10^6 m.)

A force of 200 N is applied to a body and its velocity changes from 5m/s to 10m/s in a second. What is the mass of the body?

A trolley of mass 200 kg moves with a uniform speed of 36 km/ h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s^-1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

The radius of the Moon is frac(1)(4)^th the radius of the earth and its mass is frac(1)(96)^th the mass of the earth. If the escape velocity from the surface of the earth is 11.2k ms^(-1) , find the escape velocity from the surface of the Moon.

If the radius of earth strinks by one percent it mass remaining the same by what percent will the acceleration due to gravity change at surface.

The earth’s radius is R and acceleration due to gravity as its surface is g. If a body of mass m is sent to a height of R/5 from the earth’s surface the PE increase (h = R/5) by: