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g(e) and g(p) denote the acceleration du...

`g_(e)` and `g_(p)` denote the acceleration due to gravity on the surface of the earth and another planet whose mass and radius are twice as that of earth. Then

A

`g_(p)=g_(e)`

B

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

C

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

D

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

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The correct Answer is:
To solve the problem, we need to find the relationship between the acceleration due to gravity on the surface of the Earth (denoted as \( g_e \)) and the acceleration due to gravity on the surface of another planet (denoted as \( g_p \)) whose mass and radius are both twice that of the Earth. ### Step-by-Step Solution: 1. **Understanding the Formula for Acceleration due to Gravity**: The acceleration due to gravity \( g \) on the surface of a planet is given by the formula: \[ g = \frac{G M}{R^2} \] where \( G \) is the universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. 2. **Calculate \( g_e \) for Earth**: For Earth, we denote its mass as \( M_e \) and its radius as \( R_e \). Thus, the acceleration due to gravity on Earth is: \[ g_e = \frac{G M_e}{R_e^2} \] 3. **Determine the Mass and Radius of the New Planet**: The problem states that the mass and radius of the new planet are both twice that of Earth: \[ M_p = 2 M_e \] \[ R_p = 2 R_e \] 4. **Calculate \( g_p \) for the New Planet**: Now, we can find the acceleration due to gravity on the surface of the new planet: \[ g_p = \frac{G M_p}{R_p^2} \] Substituting the values of \( M_p \) and \( R_p \): \[ g_p = \frac{G (2 M_e)}{(2 R_e)^2} \] Simplifying the denominator: \[ g_p = \frac{G (2 M_e)}{4 R_e^2} \] 5. **Relate \( g_p \) to \( g_e \)**: Now we can express \( g_p \) in terms of \( g_e \): \[ g_p = \frac{2 G M_e}{4 R_e^2} = \frac{1}{2} \cdot \frac{G M_e}{R_e^2} = \frac{1}{2} g_e \] 6. **Final Relationship**: Therefore, the relationship between the acceleration due to gravity on the new planet and that on Earth is: \[ g_p = \frac{1}{2} g_e \] ### Conclusion: The acceleration due to gravity on the surface of the new planet is half that of the acceleration due to gravity on the surface of the Earth.
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