Find the height from the surface of the moon where the value of -g- is equal to the value of -g- at a height of 57-600 km from the surface of the earth- -Take the mass of the earth- ME - 6 x 1024 kg - Mass of the moon- Mm - 7-3 x 1022 kg- radius of the earth- RE - 6400 km and radius of the moon- Rm - 1740 km-Kindly do not provide any link or hint for this qu-estion-

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Standard IX

Physics

Universal Law of Gravitation

Find the heig...

Question

Find the height from the surface of the moon where the value of 'g' is equal to the value of 'g' at a height of 57,600 km from the surface of the earth. (Take the mass of the earth, M_E = 6 x 10²⁴ kg , Mass of the moon, M_m = 7.3 x 10²² kg, radius of the earth, R_E = 6400 km and radius of the moon, R_m = 1740 km)
Kindly do not provide any link or hint for this qu/estion.

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Solution

Dear Student

$T h e v a l u e o f a c c e l e r a t i o n d u e t o g r a v i t y a t h e i g h t h i s g i v e n a s g = \frac{G M}{{(R + h)}^{2}} L e t t h e v a l u e o f g i s e q u a l a t h e i g h t h_{E} f r o m e a r t h' s s u r f a c e a n d h_{M} f r o m t h e s u r a f c e o f m o o n . g = \frac{G M_{E}}{{(R_{E} + h_{E})}^{2}} = \frac{G M_{M}}{{(R_{M} + h_{M})}^{2}} P u t t i n g g i v e n v a l u e s \frac{6 \times 10^{24}}{{(6400 + 57600)}^{2}} = \frac{7.3 \times 10^{22}}{{(1740 + h_{M})}^{2}} {(1740 + h_{M})}^{2} = {(6400 + 57600)}^{2} (\frac{7.3 \times 10^{22}}{6 \times 10^{24}}) {(1740 + h_{M})}^{2} = 49834666.6666 h_{M} = 5319.36 k m$
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