Question

Class 11th !

A body weighs 64 N on the surface of earth . What is the gravitational force on it due to the earth at a height equal to one third of the radius of the earth

No Spam !​

Answer 1

$\huge\bigstar\underline\blue{Answer}$

★ w = mg = 64N

where,

g' =$\frac{g}{1 \ + \frac{h}{re} }$

g' =$g \ + \frac{ \frac{re}{3} }{re}$

=> g'=$\frac{g}{1 \ + \frac{1}{3} }$

=>. g' = $\frac{3g}{4}$

now,

w' = mg'

=> w' = m × 3g /4

=> w' = 64 × 3/4

=> w' = 48N

Happy to help!

☺️