Question

★ Gravitational acceleration on the surface of a planet is (√6/11)g where g is the gravitational acceleration on the surface of the Earth. The average mass density of the planet is 2/3 times that if the Earth . If the escape speed on the surface of the Earth is taken to be 11km/s , what will be the escape speed on the surface of the planet in km/s ?
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Answer 1

Answer:

Solution: The acceleration due to gravity on the surface of a planet of mass M, and radius R, is given by,

Op = GMp/R.

The mass is related to density pp by Mp (4/3)TRPp. Substitute in above equation and simplify to get,

Rp = 39,/(47GP)).

The escape velocity from the surface of the planet is given by

Vp = 2GM/R = 29R, = 39/(2TGpp).

Substitute values in above equation to get,

Up

Ve

V Pp/pe

6/121 2/3

3

11

which gives Up = 3 km/s.

Answer 2

Answer:

let the gravitational acceleration on the surface be g and density bed

11'p

3

3

11

R 36 Hence, R 22 R'

R d

OX Vese

Vese = 3 km/s.