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Question: A man weighs \[50{\text{ }}kg\] at the Earth's surface. At what height above the Earth's surface doe...

A man weighs 50 kg50{\text{ }}kg at the Earth's surface. At what height above the Earth's surface does his weight become half (radius of the earth = 6400 km = {\text{ }}6400{\text{ }}km )
A. 2526 km2526{\text{ }}km
B. 6400 km6400{\text{ }}km
C. 2650 km2650{\text{ }}km
D. 3200 km3200{\text{ }}km

Explanation

Solution

To solve the provided issue, we must first ascertain the mass of man when he weighs 50N50N , but because the mass does not change, the weight reduction is due to the lack of gravity. As a result, we will determine the height ′h′'h' at which the weight is half and gravity is at that height g′g' .

Formula used:
g′=g(RR+h)2g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2}
Where, RR is the Earth's radius. R=6.4×106 mR = 6.4 \times {10^6}\,m and gg is gravity on the earth's surface, with a value of 9.81 m⋅s−29.81\,m \cdot {s^{ - 2}}.

Complete step by step answer:
When objects are close to the earth, the total mass of all particles equals the mass of the earth, and the distance is measured from the earth's centre. The weight of an object decreases with altitude because the gravitational constant (g)\left( g \right) is proportional to the square of the distance from the earth's centre.

To begin with, kilograms aren't the unit of weight; Newton is. The kilogram is the unit of mass.You must first determine the mass of man.
W=mgW = mg , as you may be aware.
As a result,
m=W1gm = \dfrac{{{W_1}}}{g} , and
⇒W1=50N.\Rightarrow {W_1} = 50N.
The mass does not vary. The lack of gravity is responsible for the weight loss.Let's say a man's weight is 25N25N at height, and gravity is g′g' at that height.Because mass does not change with height, the mass at that height would be the same. As a result, g′g' can be calculated.
g′=W2mg'= \dfrac{W_2}{m}
Where, W2=25N.{W_2} = 25N.

At that height, you now have gravity. You'll need to figure out height.
g′=g(RR+h)2g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2}
So you can easily compute the height using that formula.
g′=g(RR+h)2g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2}
W2⋅gW1=g(RR+h)2 ⇒2550=(64006400+h)2 ⇒12=64006400+h ⇒6400+h=6400×2 ⇒6400+h=9050 ⇒h=9050−6400 ∴h=2650\dfrac{{{W_2} \cdot g}}{{{W_1}}} = g{\left( {\dfrac{R}{{R + h}}} \right)^2} \\\ \Rightarrow \dfrac{{25}}{{50}} = {\left( {\dfrac{{6400}}{{6400 + h}}} \right)^2} \\\ \Rightarrow \dfrac{1}{{\sqrt 2 }} = \dfrac{{6400}}{{6400 + h}} \\\ \Rightarrow 6400 + h = 6400 \times \sqrt 2 \\\ \Rightarrow 6400 + h = 9050 \\\ \Rightarrow h = 9050 - 6400 \\\ \therefore h = 2650
At the height of 2650 km2650{\text{ }}km above the Earth's surface does his weight become half.

So, the correct option is C.

Note: When you lift an object, you are separating its distance from the Earth's centre. Because there is more distance, you have more potential energy because it would travel more distance and have more kinetic energy at the bottom of its fall if you dropped it now.