A Satellite is revolving around a planet having mass M - 8 - 1022 kg and radius R - 2 - 106 m as shown in figure- Find the number of revolutions made by the satellite around the planet in 24 hours-

The correct option is C 11 T = 2π√(R^3GM) = 2π√(8 × 10^18203× 10^ 11× 8 × 10^22) = 7800 sec .Number of revolutions = 24 × 36007800 = 11.07 ≈ 11 ...

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Standard XII

Physics

Satellites

A Satellite i...

Question

A Satellite is revolving around a planet having mass $M = 8 \times 10^{22} k g$ and radius $R = 2 \times 10^{6} m$ as shown in figure. Find the number of revolutions made by the satellite around the planet in $24$ hours.

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Solution

The correct option is C $11$
$T = 2 π \sqrt{\frac{R^{3}}{G M}} = 2 π \sqrt{\frac{8 \times 10^{18}}{\frac{20}{3} \times 10^{- 11} \times 8 \times 10^{22}}} = 7800 s e c$ .
Number of revolutions $= \frac{24 \times 3600}{7800} = 11.07 \approx 11$

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A Satellite is revolving around a planet having mass M=8×1022kg and radius R=2×106m as shown in figure. Find the number of revolutions made by the satellite around the planet in 24 hours.

The correct option is C 11T=2π√R3GM=2π  ⎷8×1018203×10−11×8×1022=7800sec.Number of revolutions =24×36007800=11.07≈11

A Satellite is revolving around a planet having mass $M = 8 \times 10^{22} k g$ and radius $R = 2 \times 10^{6} m$ as shown in figure. Find the number of revolutions made by the satellite around the planet in $24$ hours.

The correct option is C $11$
$T = 2 π \sqrt{\frac{R^{3}}{G M}} = 2 π \sqrt{\frac{8 \times 10^{18}}{\frac{20}{3} \times 10^{- 11} \times 8 \times 10^{22}}} = 7800 s e c$ .
Number of revolutions $= \frac{24 \times 3600}{7800} = 11.07 \approx 11$