The minimum projection velocity of a body from the earth's surface so that it becomes the satellite of the earth `(R_(e)=6.4xx10^(6) m)`.

Find the escape velocity of a body from the surface of the earth. Given radius of earth = 6.38 xx 10^(6) m .

At what height from the surface of earth will the value of g becomes 40% from the value at the surface of earth. Take radius of the earth = 6.4 xx 10^(6) m .

Define orbital velocity of a satellite. Derive expressions for the orbital velocity of a satellite. Show that the escape velocity of a body from the earth's surface is sqrt2 times its velocity in a circular orbit just above the earth's surface.

Find the percentage decrease in the acceleration due to gravity when a body is taken from the surface of earth of a height of 64 km from its surface [ Take R_(e) = 6.4 xx10^(6) m ]

Find the percentage decrease in the acceleration due to gravity when a body is taken from the surface of earth of a height of 64 km from its surface [ Take R_(e) = 6.4 xx10^(6) m ]

The escape velocity of a body from earth's surface is v_e . The escape velocity of the same body from a height equal to 7R from earth's surface will be

Find the percentage decrease in the acceleration due to gravity when a body is take from the surface of earth to a height of 32 km from its surface. ["Take" R_(e )=6.4xx10^(6)m]

Find the percentage decrease in the acceleration due to gravity when a body is take from the surface of earth to a height of 32 km from its surface. ["Take" R_(e )=6.4xx10^(6)m]