Question

A planet has a mass about 16 times the Earth's mass, and its radius is equal to about 4.5 Earth's radius.

(a) By setting up ratios with the corresponding Earth values, find the acceleration due to gravity at the surface of the planet.

(b) Ignoring the rotation of the planet, find the minimum escape speed from the planet.

Answer 1

Acceleration due to gravity,g= GM/R²

Acceleration due to gravity of earth,g'=GM'/R'²

Where M'=mass of earth and R'=radius of earth.

As G is constant we have g propotional to M/R²

Comparing the two we have

g/g' =R'².M/R².M'

As R=4.5R' and M=16M'

g/g'=R'².16M'/(4.5R')²M'

g/g'=16/20.25

g/g'=0.79

g=g'(0.79).

Acceleration due to gravity of planet is 0.79 times acceleration due to gravity of earth.

Or g= 0.79*9.8=7.743 m/s².

B) ;escape velocity, Ve=(2GM/R) ¹/².

As G and 2 are constants therefore Ve is proportional to (M/R)¹/².

Let V'e be the escape velocity of the earth.

Ve/V'e =(M.R'/M'.R)¹/²

Ve/V'e =(16M'. R'/M'. 4.5R')¹/².

Ve/V'e =(16/4.5)¹/²

Ve/V'e =(3. 55)¹/²

Ve/V'e =1.88

Ve =1.88V'e or 1.88 times the escape velocity of the earth.

As V'e or escape velocity of the earth=11.2 km/s

Therefore escape velocity of the planet=11.2 *1.88=21.10 km/s.