Consider a heavenly body that has mass 2M e and radius 2R e , where M e and R e are the mass and the radius of the earth respectively. The  acceleration due to gravity at the surface of this heavenly body is 
(a) 2.45 m/s2                       (b) 4.9 m/s2                 (c) 9.8 m/s2                (d) 19.6 ​m/s2

Dear Student ,
Here in this case we know that the acceleration due to gravity on the surface of earth is ,
ge=GMeRe2=9·8 m/s2Now when the the mass is M=2Me and radius is R=2Re then the acceleration due to gravity on the surface of the heavenly body is ,g'=G×2Me4Re2=12GMeRe2=ge2=9·82=4·9 m/s2
So b is the correct answer .
Regards
 

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(19.6) d
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