Question #61455

1 Answer
Oct 19, 2016

Answer:

(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#