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Given the gravitational acceleration on the surface of the moon is 1/6 of that of the...

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yoyoyeung | Student, Grade 10 | eNotes Newbie

Posted April 26, 2012 at 2:02 PM via web

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Given the gravitational acceleration on the surface of the moon is 1/6 of that of the earth and the period of the moon orbiting around the earth is 27.3 days. The mass of the moon is 7.4x10^22kg. Calculate  the radius of the moon

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mwmovr40 | College Teacher | (Level 1) Associate Educator

Posted April 27, 2012 at 2:31 AM (Answer #1)

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Along with the Three Laws of Motion for which Newton is famous, he also developed the Universal Law of Gravitation.  According to the Universal Law of Gravitation, the force of attraction between two objects with mass is given by

Fg = -G M1M2/r^2  where G is the universal gravitational constant, M1 and M2 are the two masses attracting each other, and r is the distance of separation between their centers.  If we compare this to the equation for the weight of an object W = Mg where M is the object being attracted we can see that at the surface of the Moon

Fg = W 

-GM1M/r^2  = Mg  or

g = -GM1/r^2, we can now solve this for r which will be the radius of the Moon

r =sqrt(-GM1/g), plugging the given information into this relationship we get

r =sqrt(-6.67x10^-11x7.4x10^22/-(1/6x9.80)

r = 1.76x10^6m 

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