Show that the acceleration due to gravity at the surface of moon is about 1/6 of that at the surface of the earth.
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The mass of the Moon is 7.35 x 10^22 kg and the average radius of the Moon is 1.74 x 10^6 m.
Plugging into the formula, we have:
g = (6.67 x 10^-11 N-m^2/kg^2)(7.35 x 10^22 kg )/(1.74 x 10^6 m)^2
= (4.9 x 10^12 )/(3.03 x 10^12)
= 1.6 m/s^2.
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Hey!!!!
___&_____
Here it is....
The mass of the Moon is 7.35 x 10^22 kg and the average radius of the Moon is 1.74 x 10^6 m.
Plugging into the formula, we have:
g = (6.67 x 10^-11 N-m^2/kg^2)(7.35 x 10^22 kg )/(1.74 x 10^6 m)^2
= (4.9 x 10^12 )/(3.03 x 10^12)
= 1.6 m/s^2.
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