Question

If the radius of the earth is to be increased by a factor of $5$ then by what factor its density be changed to keep the value of $g$ the same ?

• A. $\frac{1}{25}$
• B. $\frac{1}{5}$
• C. $\frac{1}{\sqrt{5}}$
• D. $5$

Answer 1

The correct option is B $\frac{1}{5}$

Acceleration due to gravity of earth is given by $$g=\frac{G{M}_e}{R_e^2}$$When the distance is increased by a factor of $5$ , then the acceleration due to gravity becomes $$g^{\prime }=\frac{G{M}_e}{(5R_e{)}^2}$$According to the data given in the question , $g=g^{\prime }$

$\therefore \frac{G{M}_e}{R_e^2}=\frac{G{M}_e}{(5R_e{)}^2}$

$\frac{\frac{4}{3}\pi R_e^3\rho }{R_e^2}=\frac{\frac{4}{3}\pi (5R_e{)}^3{\rho }^{\prime }}{(5R_e{)}^2}$

$\Rightarrow \rho =5{\rho }^{\prime }$

$\Rightarrow {\rho }^{\prime }=\frac{\rho }{5}$

Hence, option (b) is the correct answer.