Satellites
Trending Questions
- 6.4×109 J
- 4.8×109 J
- 2.7×109 J
- 3.9×109 J
- Zero
- Equal to that on the earth
- More than that on the earth
- Less than that on the earth
A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes 5.26× 103 seconds to complete a revolution while its centripetal acceleration is 9.92 m/s2 . Height of satellite above surface of earth is (Radius of earth 6.37×106 m)
70 km
120 km
170 km
220 km
- vrBrA√1.2
- vrA1.2rB
- vrB1.2rA
- vrArB√1.2
A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth; R being the radius of the earth. What will be the time period of another satellite at a height 2.5 R from the surface of the earth?
6√2 hours
6√2.5 hours
6√3 hours
12 hours
Select the correct statement from the following
The orbital velocity of a satellite increases with the radius of the orbit
Escape velocity of a particle from the surface of the earth depends on the speed with which it is fired
The time period of a satellite does not depend on the radius of the orbit
The orbital velocity is inversely proportional to the square root of the radius of the orbit
- 2
- √2
- 1√2
- √3
- 6.4×109 J
- 4.8×109 J
- 2.7×109 J
- 3.9×109 J
- Gravitational potential energy
- Angular speed
- Linear orbital speed
- Centipetal acceleration
A geostationary satellite orbits around the earth in a circular orbit of radius 36, 000 km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface (Re = 6400 km) will approximately be
12h
1 h
2 h
4 h
- Zero
- 2 sec
- 3 sec
- Infinite
Potential energy of a satellite having mass ‘m’ and rotating at a height of 6.4×106 from the earth centre is
−0.5 mgRe
−mgRe
−2 mgRe
4 mgRe
A satellite is moving around the with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1%, its speed will
Increase by 1%
Increase by 0.5%
Decrease by 1%
Decrease by 0.5%
A satellite in a force free space sweeps stationary interplanetary dust at a rate dMdt=αv, where M is the mass and v is the velocity of the satellite and α is a constant. The acceleration of the satellite is
−2αvM
−αv2M
+αv2M
−αv2
- 49GMa2
- 825GMa2
- 416GMa2
- Zero
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth :
The acceleration of S always directed towards the centre of the earth
The angular momentum of S about the cente of the earth changes in direction, but its magnitude remain constant
The total mechanical energy of S varies periodically with time
The linear momentum of S remains constant in magnitude
An earth satellite of mass m revolves in a circular orbit at a height h from the surface of the earth. R is the radius of the earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by
gR2R+h
gR
gRR+h
√gR2R+h
- −E0
- 1.5E0
- 2E0
- E0
A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes 5.26× 103 seconds to complete a revolution while its centripetal acceleration is 9.92 m/s2 . Height of satellite above surface of earth is (Radius of earth 6.37×106 m)
70 km
120 km
170 km
220 km
If G is the universal gravitational constant and ρ is the uniform density of a spherical planet. Then shortest possible period of rotation around a planet can be
√πG2ρ
√3πGρ
√π6Gρ
√3πGρ
- 4π√Rg
- 4π√2Rg
- 8π√Rg
- 8π√2Rg
- 6×108 J
- 3×108 J
- 2.614×109 J
- 5.3×109 J
- 4 days
- 8 days
- 16 days
- 32 days
A satellite is moving around the with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1%, its speed will
Increase by 1%
Increase by 0.5%
Decrease by 1%
Decrease by 0.5%
- 10−2 m
- 100 m
- 10−9 m
- 10−6 m
- R2gω
- R2ω2g
- Rgω2
- R2gω2
- f1<f2
- f1>f2
- f1=f2
- Cannot say as it depends upon mass of the space craft
Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is [g is acceleration due to gravity at Earth’s surface]
2π√2Rg
4√2π√Rg
2π√Rg
8π√Rg
- −GMm3R
- −GMm2R
- Zero
- Infinite
A satellite in a force free space sweeps stationary interplanetary dust at a rate dMdt=αv, where M is the mass and v is the velocity of the satellite and α is a constant. The acceleration of the satellite is
−2αvM
−αv2M
+αv2M
−αv2