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Escape velocity on the surface of the Earth is 11.2km/s. If any object is thrown with double the escape velocity, what would be its speed at maximum distance? Neglect the presence of sun and other celestial bodies

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Solution

Escape velocity from the Earth's surface,
ve=11.2kms1
Projection velocity from the Earth,
v1=2ve
Kinetic energy of projection
K1=12mv21=12m(2ve)2=2mv2e

or K1=12m4v2e
and gravitational potential energy
U1=GMmR=12m2GMR

or U1=12mv2e (because ve=2GMr)
If the velocity at infinity be v,
K2=12mv2 and U2=0

Now applying the principle of conservation of energy,
Total energy on Earth's surface = Total energy at infinity
or K1+U1=K2+U2

or 12m4v2e12mv2e=12mv2+0

or 12mv2e(41)=12mv2

or v2e×3=v2

or v=ve3=11.2×1.732

or v=19.398
or v=19.4kms1

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