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+2 votes
52.8k views
in Physics by (34.4k points)
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Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively. If TA and TB are the time periods of A and B respectively then the value of TB – T:

[Given : radius of earth = 6400 km, mass of earth = 6 × 1024 kg]

(1) 1.33 × 103 s

(2) 3.33 × 102 s

(3) 4.24 × 103 s

(4) 4.24 × 102 s

2 Answers

+2 votes
by (15.0k points)
selected by
 
Best answer

Correct option is (1) 1.33 × 103 s

Given: Mass of the satellite A = 200 kg

Mass of the satellite B = 400 kg

Height of satellite A = 600 Km 

Height of the satellite B = 1600 Km

The radius of satellite rA = 6400 + 600 

= 7000 Km

= 7000 × 103 m

and the radius of satellite rB = 6400 + 1600

= 8000 Km

= 8000 ×103 m

Now, by using equation (1) we have;

The time period of A, \({T_A}^2 = \frac{4\pi^2 {r_A}^3}{GM_e}\)

⇒ \({T_A} =\sqrt{ \frac{4\pi^2 {r_A}^3}{GM_e}}\)

Now, on putting the values we have;

\(T_A = \sqrt{\frac{4\pi^2(7000\times 10^{3})^3}{GM_e}}\)

and the time period of B, \({T_B}^2 = \frac{4\pi^2 {r_B}^3}{GM_e}\)

⇒ \({T_B} =\sqrt{ \frac{4\pi^2 {r_B}^3}{GM_e}}\)

Now, on putting the values we have;

\(T_B = \sqrt{\frac{4\pi^2(8000\times 10^{3})^3}{GM_e}}\)

Now, 

TB - TA = \(\sqrt{\frac{4\pi^2(8000\times 10^{3})^3}{GM_e}} - \sqrt{\frac{4\pi^2(7000\times 10^{3})^3}{GM_e}} \)

⇒ TB - TA = \(2\pi\sqrt{\frac{(8000\times 10^{3})^3}{GM_e}} - \sqrt{\frac{(7000\times 10^{3})^3}{GM_e}} \)

⇒ TB - TA = \(2\pi\sqrt{\frac{(8000 \times 10^3)^3 - (7000 \times 10^3)^3}{GM_e}}\)

⇒ TB - TA = \(2\times 3.14\sqrt{\frac{(8000 \times 10^3)^3 - (7000 \times 10^3)^3}{6.674 \times 10^{-11} \times 6 \times 10^{24}}}\)

⇒ TB - TA = 1.33 × 103 s

+3 votes
by (34.9k points)

Correct option is (1) 1.33 × 103 s

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