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The value of acceleration due to gravity...

The value of acceleration due to gravity 'g' on the surface of a planet with radius double that of the earth and same mean density as that of the earth is (`g_(e)` =acceleration due to gravity on the surface of the earth )

A

`g_(p)=2g_(e)`

B

`g_(p)=g_(e)//2`

C

`g_(p)=g_(e)//4`

D

`g_(p)=4g_(e)`

Text Solution

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The correct Answer is:
To find the value of acceleration due to gravity 'g' on the surface of a planet with a radius double that of the Earth and the same mean density as that of the Earth, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Formula for Gravitational Acceleration**: The gravitational acceleration \( g \) at the surface of a planet is given by the formula: \[ g = \frac{G \cdot M}{R^2} \] where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. 2. **Express Mass in Terms of Density and Volume**: The mass \( M \) of the planet can be expressed in terms of its density \( \rho \) and volume \( V \): \[ M = \rho \cdot V \] For a sphere, the volume \( V \) is given by: \[ V = \frac{4}{3} \pi R^3 \] Therefore, the mass becomes: \[ M = \rho \cdot \frac{4}{3} \pi R^3 \] 3. **Substitute Mass into the Gravitational Acceleration Formula**: Substituting the expression for mass into the formula for gravitational acceleration gives: \[ g = \frac{G \cdot \left(\rho \cdot \frac{4}{3} \pi R^3\right)}{R^2} \] Simplifying this, we get: \[ g = \frac{4}{3} \pi G \rho R \] 4. **Calculate for the Earth**: For the Earth, we denote the gravitational acceleration as \( g_e \): \[ g_e = \frac{4}{3} \pi G \rho R_e \] where \( R_e \) is the radius of the Earth. 5. **Calculate for the New Planet**: For the new planet, which has a radius \( R_p = 2R_e \) and the same density \( \rho \): \[ g_p = \frac{4}{3} \pi G \rho R_p = \frac{4}{3} \pi G \rho (2R_e) \] This simplifies to: \[ g_p = 2 \cdot \frac{4}{3} \pi G \rho R_e = 2g_e \] 6. **Final Result**: Therefore, the acceleration due to gravity on the surface of the planet is: \[ g_p = 2g_e \] ### Conclusion: The value of acceleration due to gravity \( g \) on the surface of the planet is \( 2g_e \), where \( g_e \) is the acceleration due to gravity on the surface of the Earth.

To find the value of acceleration due to gravity 'g' on the surface of a planet with a radius double that of the Earth and the same mean density as that of the Earth, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Formula for Gravitational Acceleration**: The gravitational acceleration \( g \) at the surface of a planet is given by the formula: \[ g = \frac{G \cdot M}{R^2} ...
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Knowledge Check

  • The value of acceleration due to gravity 'g' on the surface of the moon with radius 1/2 that of the earth and same mean density as that of the earth

    A
    g/4
    B
    g/2
    C
    4 g
    D
    8 g
  • The value of acceleration due to gravity at the surface of earth

    A
    is maximum at the poles
    B
    is maximum at the equator
    C
    remains constant everywhere on the surface of the earth
    D
    is maximum at the international time line
  • The acceleration due to gravity on the surface of earth varies

    A
    directly with longitude
    B
    directly with latitude
    C
    inversely with longitude
    D
    inversely with latitude
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