The weight of an object on the surface of the Earth is 40 N. Its weight at a height equal to the radius of the Earth is

Options

A. `sqrt(2.5) m`
B. 1 m
C. 3 mg
D. `1.5` m

Answer

A

Explanation

`v=sqrt(2a_(t)s)=sqrt(2a_(t)(piR))`

`therefore" "a_(n)=(v^(2))/(R)=2pia_(t)`
or `(a_(n))/(a_(t))=2pirArrtantheta=(a_(n))/(a_(t))=2pi`
`therefore" "theta=tan^(-1)(2pi)`