Question

The weight of a body on the surface of the earth is 100 kg. The same at a height equal to the radius of the
earth will be

Answer 1

Given:

The weight of the body on the surface of the earth = 100 kg

The height above the surface of the earth is equal to the radius of the earth i.e., h = R

To find:

The weight of the body at the height equal to the radius of the earth

Solution:

According to the Newton's Law of Gravitation, we have

$\boxed{F = \frac{GMm}{R^2}}$

where

F = gravitational force

G = gravitational constant

M = mass of earth

m = mass of body

R = distance between the masses

Also, according to the Newton's 2nd Law, we have

$\boxed{F = mg}$

where

m = mass

g = acceleration due to gravity

Comparing the above two laws of Newton, we get

∴ mg = $\frac{GMm}{R^2}$

⇒ g = $\frac{GM}{R^2}$ ....... (i)

At a height "h" above the surface of the earth

g' = $\frac{GM}{[R\:+\:h]^2}$ ....... (ii)

where

g' = acceleration due to gravity at height "h"

[R + h] = distance between the masses at height "h"

h = R [given]

On dividing equation (i) & (ii), we get

$\frac{g'}{g} = \frac{\frac{GM}{[R+h]^2}}{\frac{GM}{R^2} }$

⇒ $\frac{g'}{g} = \frac{\frac{1}{[R+h]^2}}{\frac{1}R^2} }$

⇒ $\frac{g'}{g} = \frac{R^2}{[R+h]^2}$

⇒ $\frac{g'}{g} = \frac{R^2}{[R+R]^2} ........ [h = R\: (given)]$

⇒ $\frac{g'}{g} = \frac{R^2}{[2R]^2}$

⇒ $\frac{g'}{g} = \frac{R^2}{4R^2}$

⇒ $\frac{g'}{g} = \frac{1}{4}$

⇒ g' = $\frac{g}{4}$

substituting g = 9.8 m/s²

⇒ g' = 2.45 m/s²

• We know that the mass of the body will not change in any any situation.

• The mass of the body at height h will be same as that on the surface of the earth i.e., 100 kg

• The weight of the body is the Gravitational force of attraction towards the center of the earth.

Therefore,

The weight of the body at height "h" will be given by,

= mass × acceleration due to gravity

= m × g'

= 100 × 2.45

= 245 N

Thus, the weight at a height equal to the radius of the earth will be 245 N.

---------------------------------------------------------------------------------

Also View:

The mass of body on surface of the earth is 100 kg. What will be its (i) mass and (ii) weight at an altitude of 1000 km ?

https://brainly.in/question/6965850

A body weighs 90 kg on the surface of earth how much will it weigh on surface of mars whose mass is 1/9 ?

https://brainly.in/question/1426232

The mass of the body on moon is 40kg , what is the weight on the earth ?

https://brainly.in/question/1754978

Answer 2

The same at a height equal to the radius of the  earth is 245 N

Explanation:

From question, the mass of the body = 100 kg

The weight of the body = 100 × 9.8 = 980 N

Acceleration due to gravity at height 'h' from the Earths surface is given by the formula:

g' = g/(1 + h/R)²

From question, h = R

⇒ g' = g/(1 + R/R)² = g/(1 + 1)² = g/4

Now, the weight of the body is given by the formula:

W' = mg'

W' = m × g/4

W' = 100 × 9.8/4

∴ W' = 245 N