A satellite of mass m revolves around the earth of the radius R at a height x from its surface. It g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is :
gx
An infinity number of point masses each equal to m are placed at x = 1, x = 2, x = 4, x = 8, ..... What is the total gravitational potential at x - 0?
-Gm
-2Gm
-4Gm
-8 Gm
The escape velocity of a body depends upon mass as :
m2
m3
m0
m1
A satellite moves eastwards very near the surface of the earth in the equatorial plane of the earth with speed v0. Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If R is the radius of the earth and ω be its angular speed about its own axis, then the difference in the two time period as observed on the earth will be approximately equal to :
The escape velocity for a body projected vertically upwards from the surface of earth is 11km/s. If the body is projected at an angle of 45° with the vertical, the escape velocity will be :
11km/s
11/√2 m/s
11√2 km/s
22 km/s
If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero, then the satellite will :
become stationary in its orbit
move towards the earth
continue to move in its orbit with same velocity
move tangentially to the original orbit with the same velocity
The kinetic energy needed to project a body of mass m from the earth's surface (radius R) to infinity is :
mgR
2mgR
The radius of earth is 6400 km and the value of g is 10m/s2 . If the weight of 5km body on the equator becomes zero, then the angular speed of earth will be :
Energy required to move a body of mass m from an orbit of radius 2R to 3R is :
ge and gp are accelerations due to gravity on the surface of earth and a planet respectively. The radius and mass of the planet are double the radius and mass of earth. Then :
ge = gp
ge = 2gp
gp = 2ge
ge = √2gp
