Consider a heavenly body that has mass 2Me and radius 2Re, where Me and Re are the mass and the radius of the earth respectively. The acceleration due to gravity at the surface of this heavenly body
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let g be the acceleration due to gravity on the earth so g=GMe/(Re)^2
g' be the acc. due to gravity on heavenly body
as we know that the Mass and radius is twice that of the earth so g'= G2Me/(2Re)^2
g'=2GMe/4Re^2
g'=1/2×GMe/Re^2
and as we know GMe/Re^2=g
so g'=1/2×g
g'=1/2×9.8
g'=4.9m/s^2
so g' will be half of the earth
hope it helps you....
g' be the acc. due to gravity on heavenly body
as we know that the Mass and radius is twice that of the earth so g'= G2Me/(2Re)^2
g'=2GMe/4Re^2
g'=1/2×GMe/Re^2
and as we know GMe/Re^2=g
so g'=1/2×g
g'=1/2×9.8
g'=4.9m/s^2
so g' will be half of the earth
hope it helps you....
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