QJEEM22Phdc01: JEE Main 2022 Physics - Class 11 JEEM22Phdc01 Solution: The height of any point P above the surface of earth is equal…

The acceleration due to gravity (g) at a distance (r) from the center of a planet is given by:

$g = \frac{G M}{r^2}$

where:

• G is the gravitational constant,
• M is the mass of the planet,
• r is the distance from the center of the planet.

If the height h of a point P above the surface of the Earth is equal to the diameter of the Earth, then the distance r from the center of the Earth to the point P is 3 times the radius of the Earth. Substituting this into the equation for g gives:

$g' = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$

where:

• g' is the acceleration due to gravity at the point P,
• R is the radius of the Earth.

Therefore, the value of acceleration due to gravity at point P is g/9.