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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