JEE MAIN - Physics (2022 - 25th June Morning Shift - No. 5)

The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g = acceleration due to gravity at the surface of earth).
g/2
g/4
g/3
g/9

Explanation

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.

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