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- 6300 km
- 12100 km
- 17500 km
- 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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