Question

An object weighs 100 N on earth's surface. When it is to move to a point one earth radius above earth's surface, it weigh will be?

Answer 1

weight=mg

mg=100

now acceleration due to gravity at a height R above the earth surface is

g'=g(1+h/R)^(-2)

h=R

g'=g(1+R/R)^(-2)

=g*(2)^(-2)

=g/4

new weight=mg'

=mg/4

=100/4

=25 N

Hope this helps.

Answer 2

$\LARGE{\mathsf{====Answer====}}$

Given :-

Weight = mg

[Where m = mass and g = Gravitational Force]

The value of acceleration due to gravity decreases with height. It is due to the reason that the value of acceleration due to gravity is lesser at above the surface of radius than plains.

So, h = R_______________(1)

$\huge{\boxed{\boxed{\sf{g' \: = \: g(1 \: - \: {\frac{2h}{R}})}}}}$

[Put value of (1) in above equation]

We get,

g' = gR³/(R + R)² ⇒ g/4

$\huge{\mathsf{W' \: = \: {\frac{mg}{4}}}}$

_____________[Put Value of mg]

We get,

W' = 100/4

⇒ W' = 25 N

$\large{\boxed{\red{\sf{New \:Weight \: (W') \: = \: 25 \: N}}}}$