Question

derive and expression for gravitational potential energy show that the critical velocity of body revolving in a orbit very close to the surface of s planet R and mean density P is 2R G P/3​

Answer 1

Answer:

Critical Velocity (Vc )

                               The minimum velocity required to revolve in a circular orbit around a planet is called critical velocity.

Vc = √ [ ( G M ) / R ]

_ _ _ _ _  ( 1 )

Where ,

G = Gravitational Constant

M = Mass of planet

R = Radius of planet

The density ( ρ ) of a body can be define as mass enclosed per unit volume

Therefore,

ρ = M / V

i.e.

M = ρ V

Volume of spherical planet is

V = ( 4 / 3 ) π R 3

The mass of planet can be given as

M = ρ ( 4 / 3 ) π R 3

Putting this value in equation ( 1 ) we get,

Vc = √ [ ( ( 4 / 3 ) G ρ π R 3 ) / R ]

Vc = 2 √ [ ( G ρ π R 2 ) / 3 ]

Vc = 2 R √ [ ( G ρ π ) / 3 ]

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Answer 2

Answer:

We know, minimum velocity required to revolve in a circular orbit around a planet is called critical velocity

e.g., v

c

R

GM

where M is the mass of planet , R is the radius of that planet.

we know, mass = volume × density

so,

M=

3

4

πR

3

×p

v

c

R

G

3

4

πR

3

p

so, 2R

3

Gπp

hence, the critical velocity of a body revolving in circular orbit very close to the surface of a planet of radius R and mean density ρ is

3

πρG