Question Details

A space probe projected from the earth moves round the moon in a circular orbit at a distance equal to its radius Rₘₒₒₙ= R/4 where R = radius of the earth. Its rocket launcher moves in circular orbit around the earth at a distance equal to R from its surface. The ratio of the times taken for one revolution by the probe and the rocket launcher is (Mₘₒₒₙ = M/80, where M = mass of the earth))

Options

A

√3:2

B

√5:2

C

1:1

D

2:√3

Correct Answer :

√5:2

Solution :

The correct answer is √5:2.

Step-by-Step Explanation:

1. Identify the given parameters:
- Radius of the Earth = R
- Mass of the Earth = M
- Radius of the Moon = Rmoon=R4
- Mass of the Moon = Mmoon=M80

2. Determine the orbital radii:
- The space probe moves around the Moon in a circular orbit at a distance equal to its radius (meaning a height of Rmoon above the Moon's surface). Therefore, the orbital radius of the probe (rp) from the center of the Moon is:
rp=Rmoon+Rmoon=2Rmoon=2R4=R2
- The rocket launcher moves around the Earth in a circular orbit at a distance equal to R from the Earth's surface. Therefore, the orbital radius of the launcher (rL) from the center of the Earth is:
rL=R+R=2R

3. Use the formula for the time period of a circular orbit:
The time period of one revolution of a satellite orbiting a central body of mass Mc at a radius r is given by Kepler's Third Law:
T=2πr3GMc

4. Calculate the ratio of the time periods:
For the space probe:
Tp=2πrp3GMmoon=2πR/23GM/80=2π80R38GM=2π10R3GM
For the rocket launcher:
TL=2πrL3GM=2π2R3GM=2π8R3GM
Taking the ratio of Tp to TL:
TpTL=2π10R3GM2π8R3GM=108=54=52

Thus, the ratio of the times taken for one revolution by the probe and the rocket launcher is √5:2.

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