The ratio of escape velocity at earth ve to the escape velocity at a planet vp whose radius and mean density are twice as that of earth is: - Padhle.Online
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# The ratio of escape velocity at earth ve to the escape velocity at a planet vp whose radius and mean density are twice as that of earth is:

Escape velocity is an initial velocity at which, a body when thrown will leave the gravitational field of earth and never come back and its formula is

ve = √2GMe/Re

where G is the gravitational constant, Me,Re are the mass and radius of earth

Since density is given, we can write mass as the product of volume and density

Hence M=Vρ=4πR3/3 ρ (the shape of earth and planet is spherical) V is the volume and ρ is the density, now put it in the escape velocity equation,

v = √2G×4πR3×ρ/3R

= √2G×4πR2×ρ/3

Escape velocity of earth ve=√2G×4πRe2×ρe/3

Similarly escape velocity of planet vp=√2G×4πRp2×ρp/3

The ratio is ve/vp=√2G×4πRe2×ρe/3

⇒ve/v=Re/Rp √ρep

Since it is given in the question that the radius and mean density of planet is two times to that of earth Putting this value in above equation, it becomes

ve/vp =1/2√1/2

⇒ ve/vp =1/2√2

Principal, Babu Daudayal SVM, Mathura