The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, isa)R/2b)2Rc)R/3d)3RCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Acceleration due to gravity at a height “h” is given by

g’ = g (R/R+h)²

Here,

g is the acceleration due to gravity on the surface

R is the radius of the earth

As g’ is given as g/9, we get

g/9 = g(R/R+h)²

⅓ = R/(R+h)

h=2R

Answer

To find the height at which the acceleration due to gravity becomes g/9, we can start by considering the gravitational force equation:

F = (G * M * m) / r^2

Where:
F is the gravitational force between two objects,
G is the gravitational constant,
M is the mass of one object,
m is the mass of the other object, and
r is the distance between the centers of the two objects.

We can relate the acceleration due to gravity (g) to the gravitational force by dividing both sides of the equation by the mass of the object experiencing the gravitational force (m):

g = (G * M) / r^2

Since we are interested in finding the height at which the acceleration due to gravity becomes g/9, we can set up the following equation:

g/9 = (G * M) / (R + h)^2

Where:
g/9 is the desired acceleration due to gravity,
G is the gravitational constant,
M is the mass of the Earth,
R is the radius of the Earth, and
h is the height above the surface of the Earth.

We can rearrange this equation to solve for h:

(R + h)^2 = (9 * G * M) / g

Taking the square root of both sides:

R + h = sqrt((9 * G * M) / g)

Subtracting R from both sides:

h = sqrt((9 * G * M) / g) - R

Now we have an expression for the height at which the acceleration due to gravity becomes g/9 in terms of R, the radius of the Earth. Simplifying further:

h = sqrt((9 * G * M) / g) - R

Since the radius of the Earth is R, we can substitute R for sqrt((9 * G * M) / g) - R in the equation:

h = 2R

Therefore, the correct answer is option B: 2R.

Answer

To find the height at which the acceleration due to gravity becomes g/9, let's analyze the gravitational force equation.

Gravitational force (F) between two objects is given by:
F = (G * m1 * m2) / r^2

where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.

The acceleration due to gravity (g) is the force experienced by an object of mass m due to the gravitational attraction of the Earth. It is given by:
g = (G * M) / R^2

where M is the mass of the Earth and R is the radius of the Earth.

Since the acceleration due to gravity is directly proportional to the mass of the Earth and inversely proportional to the square of the radius, we can write:
g' = (G * M') / (R')^2

where g' is the new acceleration due to gravity, M' is the new mass of the Earth, and R' is the new radius of the Earth.

We are given that g' = g/9. Substituting this into the equation above, we get:
g/9 = (G * M') / (R')^2

Simplifying, we find:
(R')^2 = (9 * M' * R^2) / (G * M)

To find the height at which the acceleration due to gravity becomes g/9, we need to find the difference between the new radius (R') and the original radius (R). Let's call this difference ΔR.

ΔR = R' - R

Now, let's substitute the expression for (R')^2 into the equation for ΔR:

ΔR = sqrt((9 * M' * R^2) / (G * M)) - R

Since the new mass M' is the same as the original mass M (since we are still considering the Earth), we can simplify further:

ΔR = sqrt((9 * R^2) / G) - R

To find the height, we need to multiply ΔR by the radius of the Earth, R:

Height = R * ΔR

Substituting the expression for ΔR, we get:

Height = R * (sqrt((9 * R^2) / G) - R)

Simplifying further, we find:

Height = R * (3R / sqrt(G) - R)

Since G is a constant, the height can be written as:

Height = 3R - R^2 / sqrt(G)

Therefore, the height at which the acceleration due to gravity becomes g/9 is given by 3R - R^2 / sqrt(G), which is equivalent to option B, 2R.

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