The period of revolution of two satellites are 3 hours and 24 hours

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1. 6300 km
2. 12100 km
3. 17500 km
4. 21000 km

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10 Questions 10 Marks 10 Mins

The correct answer is option 2) i.e. 12100 km

CONCEPT:

• The time period of the satellite: It is the time taken by the satellite to complete one revolution around the Earth.
• Consider a satellite orbiting the earth at a height h from the surface of the earth of radius R.

The circumference of orbit of satellite = 2π(R+h)

• The orbital velocity of the satellite is given by,

\(\Rightarrow v_0 =\sqrt{\frac{GM}{R+h}}\)

• Time period of the satellite is, 

\(\Rightarrow T = \frac{circumference}{orbital \: velocity} = \frac{2\pi(R+h)}{\sqrt{\frac{GM}{R+h}}} = 2\pi\sqrt{\frac{(R+h)^3}{GM}}\)

CALCULATION:

Given that:

Radius of Earth, R = 6300 km

Ratio of time period = \(\frac{T_A}{T_B} = \frac{2}{3}\)

The height at which satellite B is orbiting, hB = 2hA

• Time period of the satellite is, 

\(\Rightarrow T = 2\pi\sqrt{\frac{(R+h)^3}{GM}}\)

⇒ T ∝ (R + h)3/2

\(⇒ \frac{T_A}{T_B} = \frac{(R+h_A)^{3/2}}{(R+h_B)^{3/2} }\)

\(⇒ \frac{2}{3} = ( \frac{R+h_A}{R+2h_A })^{3/2} ⇒ (\frac{2}{3})^{2/3} =\frac{(R+h_A)}{(R+2h_A) } \)

⇒ 0.76(R + 2hA) = R + hA

⇒ 0.52hA = 0.24R

⇒ 0.52hA = 0.24 × 6300

⇒ hA = 2907 km

⇒ Distance of B from the centre of Earth = R + hB = R + 2hA = 6300 + (2 × 2907) = 12114 km ≈ 12100 km

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