Question

A body weight 98N on the surface of earth. Calculate the weight of this body at a height h=R, where R=6400km is the radius of the earth. ​

Answer 1

Answer:

Weight of body on the surface of earth = 98N

We have to find weight of body at a height of h=R. Where R is the radius of earth$.$

★ Acceleration due to gravity decreases with both; increase in height and increase in depth. It also depends on the latitude of the place.

Acceleration due to gravity has maximum value at the surface of earth.

⧪ Acceleration due to gravity at a height of h from surface is given by

$\dag\:\:\underline{\boxed{\bf{\gray{g'=\dfrac{g}{\bigg(1+\dfrac{h}{R}\bigg)^2}}}}}$

where r = radius of earth

By substituting the values, we get

➝ g' = g / (1 + R/R)²

➝ g' = g / (1 + 1)²

➝ g' = g/(2)²

➝ g' = g/4

We know that weight of body is measured as the product of mass and acceleration due to gravity.

➝ W/W' = mg/mg'

➝ 98/W' = g/(g/4)

➝ 98/W' = 4

➝ W' = 98/4

➝ W' = 24.5 N

Answer 2

Answer:

$\fbox{GIVEN}$

1. W= 95 N

2. R= 6400 km

$\fbox {TO \: FIND}$

Calculate the weight of this body at a height h=R

$\fbox{SOLUTION}$

$\bold{♠Acceleration \: due \: to \: gravity \: has \: maximum value \: at \: the \: surface \: of \: earth.}$

♠️ acceleration due to gravity at a height of h from surface is given by-

$g' = \frac{g}{ ({1 + \frac{h}{r} }^{2} )}$

By substituting the given value, we get-

➝ g' = g / (1 + R/R)²

➝ g' = g / (1 + R/R)²

➝ g' = g / (1 + 1)²

➝ g' = g/(2)²

➝ g' = g/4.

weight of body is measured as the product of mass and acceleration due to gravity.

➝ W/W' = mg/mg'

➝ 98/W' = g/(g/4)

➝ 98/W' = 4

➝ W' = 98/4

➝ W' = 24.5 N