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Standard IX

Physics

Acceleration Due to Gravity

Taking moon's...

Question

Taking moon's period of revolution about the earth as $30$ days. Calculate its distance from the earth. $(G = 6.7 \times 10 N - m^{2} / k g^{2}$ and mass of earth $= 6 \times 10^{24} k g)$ .

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Solution

We know that $T^{2} = \frac{4 π^{2} R^{3}}{G M} R^{3} = \frac{G M T^{2}}{4 π^{2}}$
Given Time T=30 days=2.59 $\times 10^{6}$ s
Thus R $= \sqrt[3]{\frac{{(2.59 \times 10^{6})}^{2} \times 6 \times 10^{- 11} \times 6 \times 10^{24}}{4 \times {3.14}^{2}}} = 189.72 \times 10^{4}$ km

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Taking moon's period of revolution about the earth as 30 days. Calculate its distance from the earth. (G=6.7×10N−m2/kg2 and mass of earth =6×1024kg).

We know that T2=4π2R3GMR3=GMT24π2Given Time T=30 days=2.59 ×106 sThus R=3 ⎷(2.59×106)2×6×10−11×6×10244×3.142=189.72×104 km

Taking moon's period of revolution about the earth as $30$ days. Calculate its distance from the earth. $(G = 6.7 \times 10 N - m^{2} / k g^{2}$ and mass of earth $= 6 \times 10^{24} k g)$ .

We know that $T^{2} = \frac{4 π^{2} R^{3}}{G M} R^{3} = \frac{G M T^{2}}{4 π^{2}}$
Given Time T=30 days=2.59 $\times 10^{6}$ s
Thus R $= \sqrt[3]{\frac{{(2.59 \times 10^{6})}^{2} \times 6 \times 10^{- 11} \times 6 \times 10^{24}}{4 \times {3.14}^{2}}} = 189.72 \times 10^{4}$ km