Question

A newly discovered planet is found to have density 2/3 ρe and radius 2r , where ρe and re are the density and radius of earth, respectively. the surface gravitational field of the planet is most nearly

Answer 1

The surface gravitational field of the planet can be calculated using formula:
$g= \frac{G*m}{ r^{2} }$
Where:
G=gravitational constant
m=mass of planet
r=radius of planet

For Earth this gives:
$g= \frac{G* m_{E} }{( r_{E})^{2} } \\ \\ g= \frac{G* \rho_{E}* \frac{4}{3} * ( r_{E} )^{3}* \pi }{( r_{E})^{2}} \\ \\ g= {G* \rho_{E}* \frac{4}{3} * ( r_{E} ) * \pi }$

Before we calculate gravitational field of a planet we need to calculate it's mass. It can be calculated using density.
$\rho= \frac{m}{V} \\ m=\rho*V \\ m= \frac{2}{3} * \rho_{E} * \frac{4}{3} * (2r_E} )^{3} * \pi \\ \\ m= \frac{8}{3}* \rho_{E} * (2r_E} )^{3} * \pi$

Gravitational field of a planet is:
$g= \frac{G* \frac{8}{3}* \rho_{E} * ( 2r_{E} )^{3} * \pi }{( 2r_{E})^{2} } \\ \\g= \frac{16}{3}*G*\rho_{E}*r_{E}* \pi$

Now we compare Earth's and planet's gravitational fields:
${G* \rho_{E}* \frac{4}{3} * ( r_{E} ) * \pi }.....................\frac{16}{3}*G*\rho_{E}*r_{E}* \pi \\ \\1.....................4$

Gravitational field on a planet is 4 times greater than gravitational field on Earth.