Question:

The density of core of a planet is $\rho_1$ and that of the outer shell is $\rho_2$. The radii of core and that of the planet are $R$ and $2R$ respectively. Gravitational acceleration at the surface of planet is same as at a depth. The ratio between $\frac{\rho _{1}}{\rho _{2}}$ is

Updated On: Jul 7, 2022
  • $2.3$
  • $4.5$
  • $3.2$
  • $5.4$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

$\frac{Gm_{1}}{R^{2}} = \frac{G\left(m_{1}+m_{2}\right)}{\left(2R\right)^{2}}$ $4m_{1} = m_{1} + m_{2}$ $3m_{1} = m_{2}$ $3\left(\frac{4}{3}\pi R^{3}\rho_{1}\right)$ $ = \frac{4}{3}\pi\left[8R^{3}-R^{3}\right]\rho_{2}$ $3\rho_{1} = 7\rho_{2}$ $\frac{\rho _{1}}{\rho _{2}} = \frac{7}{3} = 2.3$
Was this answer helpful?
0
0

Concepts Used:

Gravitational Potential Energy

The work which a body needs to do, against the force of gravity, in order to bring that body into a particular space is called Gravitational potential energy. The stored is the result of the gravitational attraction of the Earth for the object. The GPE of the massive ball of a demolition machine depends on two variables - the mass of the ball and the height to which it is raised. There is a direct relation between GPE and the mass of an object. More massive objects have greater GPE. Also, there is a direct relation between GPE and the height of an object. The higher that an object is elevated, the greater the GPE. The relationship is expressed in the following manner:

PEgrav = mass x g x height

PEgrav = m x g x h

Where,

m is the mass of the object,

h is the height of the object

g is the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.