GATE-2018 (24) :Wyllie – Time average equation

CSIR-NET 2014

113. A,B and C  are three spherical shells of equal thickness and of radii 3,000km , 4,000km and 5,000km respectively, with a material of uniform density. If gA, gB and gC are the gravity values on their respective surfaces ,then

1) gA=gB=gC 

2) 3gA=4gB=5gC

3) 12gA=15gB=20gC

4) 20gA=15gB=12gC

(Special thanks to Anand, AU)

Solution:

Acceleration due to gravity $g=\frac{Gm}{r^{2}}$----(1)

From the above problem, we need a relation between 

the acceleration due to gravity and radius.

So, Density (ρ) = mass / volume ----------------(2)

mass = density * volume

$m = \rho*v$----------(3)

eq (3) substitute in eq(1)

g= ( (G* ρ* V) / r²) ----------(4)

Volume of the spherical shell (V) = ( (4/3)* Π*r³ )

g= ( ((4/3)* Π*r³* G*ρ ) / r² )

g= ( (4/3)* Π*r* G*ρ ) -------------------(5)

g_A= ( (4/3)* Π*r_A* G*ρ ) ---------------(6)

g_B= ( (4/3)* Π*r_B* G*ρ ) ----------------(7)

g_C= ( (4/3)* Π*r_C* G*ρ ) ---------------(8)

Therefore, 
 

$g_{A}:g_{B}:g_{c} = G *\frac{4}{3}\pi r_{A} *\rho: G* \frac{4}{3} \pi r_{B}*\rho:G*\frac{4}{3}\pi r_{C}*\rho$

                      $=r_{A} :r_{B}:r_{C}$

                      $= 3000:4000:5000$

$g_{A}:g_{B}:g_{C} =3:4:5$

$20g_{A}:15g_{B} :12g_{C} =60:60:60 =1:1:1$

$20g_{A} =15g_{B}=12g_{C}$