The height at which the weight of a body becomes 1/16th, its weight on the surface of earth (radius R), is : [2012]a)5Rb)15Rc)3Rd)4RCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Answer

Answer

To solve this problem, we need to use the concept of gravitational force and the inverse square law of gravity.

Gravitational force is given by the equation:

F = (G * m1 * m2) / r^2

Where:
F is the gravitational force between two objects,
G is the gravitational constant,
m1 and m2 are the masses of the two objects, and
r is the distance between the centers of the two objects.

Let's assume the mass of the body is 'm' and its weight on the surface of the Earth is 'W'. Using the formula for weight, we have:

W = m * g

Where:
g is the acceleration due to gravity on the surface of the Earth.

Now, we need to find the height at which the weight of the body becomes 1/16th of its weight on the surface of the Earth. Let's call this height 'h'.

We can calculate the gravitational force at the height 'h' using the formula:

F' = (G * m * M) / (R + h)^2

Where:
M is the mass of the Earth and
R is the radius of the Earth.

To find the height 'h', we can equate the gravitational force at the height 'h' to 1/16th of the weight on the surface of the Earth:

F' = (1/16) * W

Now, let's substitute the values and solve for 'h':

(G * m * M) / (R + h)^2 = (1/16) * (m * g)

Canceling out the mass 'm' and rearranging the equation, we get:

(G * M) / (R + h)^2 = (1/16) * g

Now, let's solve for 'h':

(R + h)^2 = (16 * G * M) / g

Taking the square root of both sides, we get:

R + h = sqrt((16 * G * M) / g)

Subtracting R from both sides, we get:

h = sqrt((16 * G * M) / g) - R

This is the expression for the height 'h' at which the weight of the body becomes 1/16th of its weight on the surface of the Earth.

Now, let's analyze the answer options:

a) 5R
b) 15R
c) 3R
d) 4R

Comparing the expression for 'h' with the answer options, we can see that h is equal to 3R. Therefore, the correct answer is option C.

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