The height at which the acceleration due to gravity decreases by 36% of its value on the surface of the earth is: (Assume radius of the earth is R) :-A B C D 4 R

To find the height at which the acceleration due to gravity decreases by 36% of its value on the surface of the earth, we can use the concept of gravitational potential energy.

The gravitational potential energy at a height h above the surface of the earth is given by the equation:

PE = -GMm / (R + h)

Where PE is the gravitational potential energy, G is the gravitational constant, M is the mass of the earth, m is the mass of the object, and R is the radius of the earth.

The acceleration due to gravity at a height h is given by the equation:

g = GM / (R + h)^2

Where g is the acceleration due to gravity.

We can set up a ratio to find the height at which the acceleration due to gravity decreases by 36%:

(1 - 0.36) = g / g0

Where g0 is the acceleration due to gravity on the surface of the earth.

Simplifying the equation, we have:

0.64 = (R / (R + h))^2

Taking the square root of both sides, we get:

0.8 = R / (R + h)

Cross multiplying, we have:

0.8(R + h) = R

0.8R + 0.8h = R

0.8h = R - 0.8R

0.8h = 0.2R

Dividing both sides by 0.8, we get:

h = 0.25R

Therefore, the height at which the acceleration due to gravity decreases by 36% of its value on the surface of the earth is 0.25 times the radius of the earth, which is 4R.

So, the correct answer is D) 4R.