Question

Question #61455

Answer 1

(b) `1/3`

Explanation:

As per Law of Universal Gravitation the force of attraction between two bodies is given as

`F_G =G (M_1.M_2)/r^2`

Where `G` is gravitational constant.
It has the value `6.67408 xx 10^-11 m^3 kg^-1 s^-2`.
Comparing it with the expression
`F("weight")=mg`
we get value of `g=G (M_e)/r_e^2` ......(1)
If `rho` is density of earth, then
`rho=M_e/(Volume)`
`rho=M_e/(4/3pir_e^3)`
`=>M_e=4/3pirhor_e^3`
Inserting above value of mass of earth in (1) we get
`g=G (4/3pirhor_e^3)/r_e^2`
`=>g=G 4/3pirhor_e`
`=>gproprhor_e`
From above we see that if radius of earth is increased by a factor of `3`, to keep value of `g` same we need to decrease value of `rho` by a factor of `3`.
Hence we get `1/3`