derivation of escape velocity.
Answers
Answered by
1
dw =fdx
=fdxcos o
=fdx
dw=GMmdx/x^2
integrate
/dw= /GMmdx/x^2
w=GMm/R
GMm/R=1/2mv^2
Ve^2=2gR
=fdxcos o
=fdx
dw=GMmdx/x^2
integrate
/dw= /GMmdx/x^2
w=GMm/R
GMm/R=1/2mv^2
Ve^2=2gR
Answered by
4
Gravitational potential energy + kinetic energy = 0
Also we know that
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