# Q-Let g be the acceleration due to gravity at earths surface and k the rotational kinetic energy of the earth.Suppose the earths radius decreases by 2%.Keeping all other quantities constant,then(a)g increases by 2% and k increases by 2%(b)g increases by 4% and k increases by 4%. Please show the solution

Akash Kumar Dutta
98 Points
9 years ago

Dear Nimish,

the angular momentum will remaine conserved. so
I(1)w(1)=I(2)w(2)  => 2/5.MR(1)^2.w(1)=2/5.MR(2)^2.w(2)
so w(1)/w(2) = [R(2)/R(1)]^2
=> w(1)/w(2) = (49/50)^2 .......eq.1                  since R(2)= R(1)-R(1)/50
now
k(1)=1/2.I(1).w(1)^2
K(2)=1/2.I(2).w(2)^2
=> k(1)/k(2) =  I(1)w(1)^2 / I(2)w(2)^2 = [ R(1)w(1)/R(2)w(2) ]^2
we know R(1)/R(20 = 50/49.....so
=> k(1)/k(2) = [ 49/50 ] ^2             {from equation (1)}
hence %increase in k = (k2 -  k1) / k1  . 100
=  (k2/k1 - 1).100
=  (99.100/49.49)
~  4%
similarly we get for g...we will get 4% increment.
Hence the ans is b(ANS)

Regards.

Abhishekh kumar sharma
34 Points
9 years ago

the angular momentum will remaine conserved. so
I(1)w(1)=I(2)w(2)  => 2/5.MR(1)^2.w(1)=2/5.MR(2)^2.w(2)
so w(1)/w(2) = [R(2)/R(1)]^2
=> w(1)/w(2) = (49/50)^2 .......eq.1                  since R(2)= R(1)-R(1)/50
now
k(1)=1/2.I(1).w(1)^2
K(2)=1/2.I(2).w(2)^2
=> k(1)/k(2) =  I(1)w(1)^2 / I(2)w(2)^2 = [ R(1)w(1)/R(2)w(2) ]^2
we know R(1)/R(20 = 50/49.....so
=> k(1)/k(2) = [ 49/50 ] ^2             {from equation (1)}
hence %increase in k = (k2 -  k1) / k1  . 100
=  (k2/k1 - 1).100
=  (99.100/49.49)
~  4%
similarly we get for g...we will get 4% increment.
Hence the ans is b(ANS)