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= (G*m)/( r^(2) )`
Where:
G=gravitational constant
m=mass of planet
r=radius of planet

For Earth this gives:

`g= (G* m_(E) )/(( r_(E))^(2) ) \\  \\ g= (G* \rho_(E)* (4)/(3) * ( r_(E) )^(3)* \pi   )/(( r_(E))^(2))  \\  \\ g= {G* \rho_(E)* (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= (m)/(V) \\ m=\rho*V \\ m= (2)/(3) * \rho_(E) * (4)/(3) * (2r_E} )^(3) * \pi \\ \\ m= (8)/(3)* \rho_(E) * (2r_E} )^(3) * \pi`

Gravitational field of a planet is:

`g= (G* (8)/(3)* \rho_(E) *  ( 2r_(E) )^(3) * \pi )/(( 2r_(E))^(2) ) \\ \\g= (16)/(3)*G*\rho_(E)*r_(E)* \pi`

Now we compare Earth's and planet's gravitational fields:

`{G* \rho_(E)* (4)/(3) * ( r_(E) ) * \pi }.....................(16)/(3)*G*\rho_(E)*r_(E)* \pi \\ \\1.....................4`

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