Home
Class 11
PHYSICS
The radius and acceleration due to gravi...

The radius and acceleration due to gravity of moon are `1/4` and `1/5` that of earth. The ratio of the mass of earth to mass of moon is

A

`1:80`

B

`80:1`

C

`1:20`

D

`20:1`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the ratio of the mass of the Earth (M) to the mass of the Moon (m) given that the radius of the Moon is \( \frac{1}{4} \) that of the Earth and the acceleration due to gravity on the Moon is \( \frac{1}{5} \) that of the Earth. ### Step-by-step Solution: 1. **Identify the given values:** - Let \( R_e \) be the radius of the Earth. - The radius of the Moon \( R_m = \frac{1}{4} R_e \). - Let \( g_e \) be the acceleration due to gravity on Earth. - The acceleration due to gravity on the Moon \( g_m = \frac{1}{5} g_e \). 2. **Use the formula for acceleration due to gravity:** The formula for the acceleration due to gravity is given by: \[ g = \frac{GM}{R^2} \] where \( G \) is the universal gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius. 3. **Write the equations for Earth and Moon:** For Earth: \[ g_e = \frac{G M}{R_e^2} \] For Moon: \[ g_m = \frac{G m}{R_m^2} \] 4. **Substituting the values for the Moon:** Substitute \( R_m = \frac{1}{4} R_e \) into the equation for \( g_m \): \[ g_m = \frac{G m}{\left(\frac{1}{4} R_e\right)^2} = \frac{G m}{\frac{1}{16} R_e^2} = \frac{16 G m}{R_e^2} \] 5. **Set up the ratio of the accelerations:** Now, we can set up the ratio of the gravitational accelerations: \[ \frac{g_e}{g_m} = \frac{\frac{G M}{R_e^2}}{\frac{16 G m}{R_e^2}} = \frac{M}{16 m} \] 6. **Substituting the known values of \( g_m \) and \( g_e \):** We know that \( g_m = \frac{1}{5} g_e \): \[ \frac{g_e}{\frac{1}{5} g_e} = \frac{M}{16 m} \] Simplifying gives: \[ 5 = \frac{M}{16 m} \] 7. **Rearranging the equation:** Rearranging gives: \[ M = 80 m \] 8. **Finding the ratio of the masses:** Therefore, the ratio of the mass of the Earth to the mass of the Moon is: \[ \frac{M}{m} = 80 \] This can be expressed as: \[ \text{Ratio of mass of Earth to mass of Moon} = 80:1 \] ### Final Answer: The ratio of the mass of Earth to the mass of Moon is \( 80:1 \). ---
Promotional Banner

Similar Questions

Explore conceptually related problems

The acceleration due to gravity on the moon is

The ratio of the radius of the earth to that of the motion is 10. the ratio of the acceleration due to gravity on the earth to that on the moon is 6. The ratio of the escape velocity from the earth's surface to that from the moon is

The acceleration due to gravity on the surface of the moon is (1)/(6)th of that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be

The acceleration due to gravity on the surface of the moon is (1)/(6)th of that on the surface of earth and the diameter of the moon is one-fourth that of earth. The ratio of escape velocities on earth and moon will be

The acceleration due to gravity on the moon is one-sixth that on earth. If the average density of moon is three-fifth that of earth, the moon's radius in terms of earth's radius R_(e) is -

The acceleration due to gravity on the moon is only one sixth that of earth.It the earth and moon are assume to have the same density, the ratio of the radii of moon and earth will be

How is the acceleration due to gravity on the surface of the earth related to its mass and radius ?

An earth satellite of mass m revolves in a circular orbit at a height h from the surface of the earth. R is the radius of the earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

Find the acceleration due to gravity of the moon at a point 1000 km above the moon's surface. The mass of the moon is 7.4 x10^22 ' kg and its radius is 1740 km.

The acceleration due to gravity on the surface of moon is 1.7 ms^(-2) . What is the time period of a simple pendulum on the moon if its time period on the earth is 3.5 s ? (g on earth = 9.8 ms^(-2) )