Weightlessness experienced while orbiting the earth in a space-ship is the result of:
1. Inertia
2. Acceleration
3. Zero gravity
4. Freefall towards the earth
The escape velocity for a rocket from the earth is \(11.2\) km/s. Its value on a planet where the acceleration due to gravity is double that on the earth and the diameter of the planet is twice that of the earth (in km/s) will be:
1. | \(11.2\) | 2. | \(5.6\) |
3. | \(22.4\) | 4. | \(53.6\) |
The escape velocity from the earth is about 11 km/second. The escape velocity from a planet having twice the radius and the same mean density as the earth is
(1) 22 km/sec (2) 11 km/sec
(3) 5.5 km/sec
(4) 15.5 km/sec
What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km
(a) rad/sac
(b) 6.4 rad/sac
(c) rad/sac
(d) 8.7 rad/sac
If g is the acceleration due to gravity at the earth's surface and r is the radius of the earth, the escape velocity for the body to escape out of the earth's gravitational field is:
(1) gr
(2)
(3) g/r
(4) r/g
The escape velocity of a projectile from the earth is approximately
(1)11.2 m/sec (2)112 km/sec
(3)11.2 km/sec (4)11200 km/sec
The escape velocity of a particle of mass m varies as:
(1)
(2) m
(3)
(4)
Acceleration due to gravity is ‘g’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km)
(1) 0.9 g
(2) 0.99g
(3) 0.8 g
(4) 1.01 g
The time period of a simple pendulum on a freely moving artificial satellite is
(1) Zero
(2) 2 sec
(3) 3 sec
(4) Infinite
For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of:
1. | \(2\) | 2. | \(\sqrt{2}\) |
3. | \(1/\sqrt{2}\) | 4. | \(4\) |