A particle is projected from point `A`, that is at a distance `4R` form the centre of the earth, with speed `V_(1)` in a direction making `30^(@)` with the line joining the centre of the earth and point `A`, as shown. Consider gravitational interaction only between thesetwo. (Use `(GM)/R=6.4xx10^(7) m^(2)//s^(2)`). The speed `V_(1)` if particle pasess grazing the surface of the earth is

A particle is projected from the surface of the earth with a speed of 20 m* s^(-1) at an angle 30^@ with the horizontal . The range of that particle is

A particle is projected from the surface of the earth with a speed of 20 m* s^(-1) at an angle 30^@ with the horizontal . The time of flight of that particle is

A particle is projected from the surface of the earth with a speed of 20 m* s^(-1) at an angle 30^@ with the horizontal . The maximum height the particle can reach is

A body at rest starts from a point at a distance r (gtR) from the centre of the Earth. If M and R stand for the speed of the body when it reaches the Earth surface is

A body at rest starts from a point at a distance r (gtR) from the centre of the Earth. If M and R stand for the speed of the body when it reaches the Earth surface is

Mass of the earth is 81 times has mass of the moon. The distance between earth and moon is 4xx10^5 km. calculate the position of the point on the line joining the centres of earth and moon where the gravitational field is zero.

An asteroid was fast approaching the earth. Scientists fired a rocket which hit the asteroid at a distance of 5 R from the centre of the earth (R = radius of the earth). mmediately after the hit the asteroid’s velocity (V_(0)) was making an angle of theta=30^(@) with the line joining the centre of the earth to the asteroid. The asteroid just grazed past the surface of the earth. Find V_(0) [Mass of the earth = M]