Q 1 :

A body weights 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is             [2025]

  • 32 N

     

  • 36 N

     

  • 16 N

     

  • 27 N

     

(4)

Given, W=48N and h=R3

Gravitational force at a height h above the surface of the Earth, F=GmM(R+h)2

or F=G(Wg)×M(R+R3)2                 [Weight, W=mg or m=Wg]

or   F=G×48×M×9g×16R2

Substituting, g=GMR2, we get

F=48×916=27N



Q 2 :

A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?        [2020]

  • 48 N

     

  • 32 N

     

  • 30 N

     

  • 24 N

     

(2)

Gravitational force at a height h,

mgh=mg0(1+hr)2=72(1+R/2R)2  or  mgh=32N

or   Fg=32N



Q 3 :

A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth?           [2019]
 

  • 100 N

     

  • 150 N

     

  • 200 N

     

  • 250 N

     

(1)

Acceleration due to gravity at a depth d,

gd=g(1-dR)

For   d=R2gd=g(1-R/2R)=g2

Required weight W'=mgd=mg2=W2=2002=100 N

 



Q 4 :

The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then      [2017]
 

  • d=1 km

     

  • d=32 km

     

  • d=2 km

     

  • d=12 km

     

(3)

The acceleration due to gravity at a height h is given as

         gh=g(1-2hRe)

where Re is the radius of the Earth.

The acceleration due to gravity at a depth d is given as

        gd=g(1-dRe)

Given,  gh=gd

  g(1-2hRe)=g(1-dRe)

or    d=2h=2×1=2km                                    (h=1km)



Q 5 :

Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by              [2016]
 

  •  

  •  

  •  

  •  

(2)

Acceleration due to gravity is given by

g={43πρGr   ;rR43πρR3Gr2  ;r>R