If `v_(0)` orbital velocity of a surface to a circular orbital close in the earth's surface and `v_(1)` scape velocity from earth then relation between the two is

Options

A. `v_(e)= 2v_(o)`
B. `v_(e)= sqrt(3v_(o))`
C. `v_(e)= v_(o)sqrt(2)`
D. `v_(o)= v_(e)`

Answer

c

Explanation

The orbital velocity of a sateline close to earth is given by
`v_(0) = sqrt(gR_(e))`….(i)
where `R_(e)` radius of earth and g is acceleration that to gravity .The escape velocity of a thrown from earth 's suface is
`v_(0) = sqrt(2gR_(e))`
Thus, `(v_(0))/(v_(e)) = sqrt(gR_(e))/(sqrt(2gR_(e))) =(1)/(sqrt(2))`
`:. v_(e) = sqrt(2 v_(e))`