A body weighs 900N on earth's surface. At a distance of twice earth radius above the earth's surface , the weight of the body is: *100N *200N *300N *450?

1 Answer
Carlos S.
Feb 2, 2018

100 N.

Explanation:

The body's mass will aways be the same. However, the acceleration of gravity, #g#, varies according to the altitude. According to Newton's gravitational law:

#F = (GmM)/d^2#,

where #G = 6.67 * 10^-11 m³kg^(-1)s^(-2)# is the gravitational constant, #m# and #M# are the masses that attract each other and #d# is the distance between them.

On the Earth's surface, #d = R_E#, where #R_E# is the Earth's radius. In this case, #F = 900 N# (the body's weight). At a distance of twice #R_E# above the Earth's surface (which equals a distance of #3R_E#), then:

#F' = (GmM)/(3R_E)^2#;

#F' = (GMm)/(9R_E^2)#.

Since #F = (GMm)/(R_E^2)# (the body's weight on the Earth's surface), then:

#F' = F/9#.

Since F = 900 N:

#F' = 900/9#;

#F' = 100 N.#

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