AP EAPCET 07-Jul-2022 Shift 1 Paper

Let, d= Density of planet,
M= Mass of planet
R= Radius of planet
then density, d=

Mass

Volume

⇒d=

M

( 4 3 πR3)

⇒R3=

3M

4πd

⇒R=(

3M

4πd

)

1

3

...(i)
Now, acceleration due to gravity g is given by
g=

GM

R2

Substituting R from Eq.(i) we gẹt,
⇒g=

GM

R2

=

GM

( 3 4π ⋅ M d ) 2 3

⇒g∝M

1

3

d

2

3

Hence, ratio of acceleration due to gravity on surface of planet and that on surface of earth will be

g′

g

=

(M 1 3 .

M 1 3

(d) 2 3

(d) 2 3

=(

M′

M

)

1

3

⋅(

d′

d

)

2

3

Given,

M′

M

=8 and

d′

d

=27
So

g′

g

=8

1

3

⋅27

2

3

=2×9
or g′=18g