A satellite in a circular orbit of radius R has a period of 4 h. Another satellite with orbital radius 3 R around the same planet will have a period (in hours)

• 16

• 4

• $4\sqrt{27}$

• $4\sqrt{8}$

The acceleration due to gravity becomes $\left(\frac{\mathrm{g}}{2}\right)$ (g = acceleration due to gravity on the surface of the earth) at a height equal to

• 4R

• $\frac{\mathrm{R}}{4}$

• 2R

• $\frac{\mathrm{R}}{2}$

A planet revolves around the sun in an elliptical orbit. The linear speed of the planet will be maximum at

• D

• B

• A

• C

74.

An astronaut on a strange planet finds that acceleration due to gravity is twice as that on the surface of earth. Which of the following could explain this ?

• Both the mass and radius of the planet are half as that of earth

• Radius of the planet is half as that of earth, but the mass is the same as that of earth

• Both the mass and radius of the planet are twice as that of earth

• Mass of the planet is half as that of earth, but radius is same as that of earth

Two satellites of mass m and 9m are orbiting a planet in orbits of radius R. Their periods of revolution will be in the ratio of

• 9 : 1

• 3 : 1

• 1 : 1

• 1 : 3

Faintest stars are called

• zero magnitude stars

• second magnitude stars

• sixth magnitude stars

• dwarfs

A planet moving around sun sweeps area A₁ in 2 days, A₂ in 3 days and A₃ in 6 days. Then, the relation between A₁, A₂ and A₃ is

• 6A₁ = 3A₂ = 2A₃

• 3A₁ = 2A₂ = A₃

• 2A₁ = 3A₂ = 6A₃

• 3A₁ = 2A₂ = 6A₃

Earth is moving around the sun in elliptical orbit as shown. The ratio of OB and OA is R. Then the ratio of Earth's velocities at A and B is

• R^-1

• $\sqrt{\mathrm{R}}$

• R

• R^2/3

What is a period of revolution of the earth satellite ? Ignore the height of satellite above the surface of the earth. Given : 

1. The value of gravitational acceleration g = 10 ms^-2
2. Radius of the earth, R_E = 6400 km (Take π = 3.14)

• 85 min

• 156 min

• 83.73 min

• 90 min

A period of geostationary satellite is

• 24 h

• 12 h

• 30 h

• 48 h