Question

calculate the escape velocity on the surface of the planet having radius 1100km and acceleration due to gravity on the surface of planet is 1.6 meter per second squares​

Answer 1

☆VALID ANSWER☆

REFER THE ATTACHEMENT

Given :   ve=11  km/s        Rp=2Re           ρp=ρe

Escape velocity    v=R2GM=R2G×34πR3ρ=38πGR2ρ

⟹    v∝ρR2

∴    vevp=ρeRe2ρpRp2

Thus we get,     11vp=ρeRe2

Answer 2

Given:-

→ Radius of the planet = 1100 km

→ Acceleration due to gravity on the

surface of the planet = 1.6 m/s²

To find:-

→ Escape velocity on the surface of

the planet.

Solution:-

Firstly, let's convert the radius of the planet from km to m.

=> 1km = 1000m

=> 1100km = 1100(1000)

=> 1100000 m

Now, we know that :-

vₚ = √(2gₚrₚ)

Where :-

• vₚ is the escape velocity of the planet.

• gₚ is the acceleration due to gravity on

the surface of the planet.

• rₚ is the radius of the planet.

Substituting values, we get :-

=> vₚ = √(2×1.6×1100000)

=> vₚ = √3520000

=> vₚ = 1876.6 m/s

=> vₚ = 1.87 km/s

Thus, escape velocity on the surface of the planet is 1.87km/s .

Some Extra Information:-

The minimum velocity that a moving body must have to escape from the gravitational field of a. celestial body is called it's escape velocity.