QJEEM23Ph80a2: JEE Main 2023 Physics - Class 11 JEEM23Ph80a2 Solution: The weight of a body at the surface of earth is 18 N. The…

The weight of an object at a height h from the Earth's surface is given by:

$$W^{\prime }=W{\left(\frac{R}{R+h}\right)}^2$$

where:

• W' is the weight at height h,
• W is the weight at the Earth's surface,
• R is the radius of the Earth,
• h is the height above the Earth's surface.

In this problem, the weight at the Earth's surface W is given as 18 N, the radius of the Earth R is given as 6400 km, and the height h is given as 3200 km. Substituting these values into the equation gives:

$$W^{\prime }=18N{\left(\frac{6400km}{6400km+3200km}\right)}^2=18N{\left(\frac{2}{3}\right)}^2=18N\cdot \frac{4}{9}=8N$$

Therefore, the weight of the body at an altitude of 3200 km above the Earth's surface is 8 N.