JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 20)
The radius in kilometer to which the present radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is ___________.
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64
Explication
$${V_{es}} = \sqrt {{{2GM} \over R}} $$
$${V_{es}}\sqrt R $$ = const
$$ \therefore $$ $${V_{es}}.\sqrt R = 10{V_{es}}\sqrt {R'} $$
$$ \Rightarrow $$ $$R' = {R \over {100}}$$ = 64 km
$${V_{es}}\sqrt R $$ = const
$$ \therefore $$ $${V_{es}}.\sqrt R = 10{V_{es}}\sqrt {R'} $$
$$ \Rightarrow $$ $$R' = {R \over {100}}$$ = 64 km