Calculation of true thickness of the bed

CSIR-NET 2015

 

 

22. In order to have the same value of the surface acceleration due to gravity as on earth, a planet twice its radius must have a mean relative density of.

 

1)2.75                      2)5.5                    3)1.4                 4)2.25

Solution:

Formula,

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

From the above problem, we need a relation between the radius and density.

So,

$Density= \frac{mass}{volume}$

$mass=Density\times volume$

$m=\rho \times v ---(2)$

Substitute equation (2) in equation (1)

$g= \frac{G\times \rho\times v}{r^2}$

then,

$\frac{G\times \rho_1\times v_1}{r_1^2}= \frac{G\times \rho_2\times v_2}{r_2^2}---(3)$

Volume of the Earth and planets are, $v=\frac{4}{3}\pi r^3$

Substitute the above volume information in equation (3)

$\frac{G\times \rho_1\times \frac{4}{3}\pi r_1^3}{r_1^2}= \frac{G\times \rho_2\times \frac{4}{3}\pi r_2^3}{r_2^2}$

Therefore simplifying the above equation we get,

$\rho_1\times r_1=\rho_2\times r_2$

$\frac{\rho_2}{\rho_1}=\frac{r_1}{r_2}---(4)$

From above problem

$r_2=2\times r_1$

$\rho_1=5.5 gm/cc$ Earth density

Substitute above values in equation (4)

$\frac{\rho_2}{\rho_1}=\frac{r_1}{2 \times r_1}$

$\frac{\rho_2}{\rho_1}=\frac{1}{2}$

$\rho_2=\frac{5.5}{2}$

$\rho_2=2.75 gm/cc$