The mass of a planet is double the mass of the earth and its radius is half of that of the earth. If F is the force of attraction on an object on the surface of earth, then the force of attraction on the same object on the surface of that planet is ________

Options

A. ` 2F `
B. ` 4F `
C. ` ( F)/(2)`
D. ` 8F `

Answer

To solve the problem, we need to determine the force of attraction on an object on the surface of a planet with specific mass and radius characteristics compared to Earth. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify Given Information:**
- Mass of the Earth (Me)
- Radius of the Earth (Re)
- Mass of the planet (Mp) = 2 * Me (double the mass of Earth)
- Radius of the planet (Rp) = Re / 2 (half the radius of Earth)

2. **Understand the Formula for Gravitational Force:**
The gravitational force \( F_g \) acting on an object of mass \( m \) at the surface of a planet is given by the formula:
\[ F_g = \frac{G \cdot M \cdot m}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

3. **Calculate the Gravitational Force on Earth:**
For Earth, the force of attraction \( F \) can be expressed as:
\[ F = \frac{G \cdot Me \cdot m}{Re^2} \]

4. **Calculate the Gravitational Force on the Planet:**
For the planet, substituting the values of \( Mp \) and \( Rp \):
\[ F_g \text{ (on planet)} = \frac{G \cdot Mp \cdot m}{Rp^2} \]
Substituting \( Mp = 2Me \) and \( Rp = \frac{Re}{2} \):
\[ F_g \text{ (on planet)} = \frac{G \cdot (2Me) \cdot m}{\left(\frac{Re}{2}\right)^2} \]

5. **Simplify the Expression:**
The denominator becomes:
\[ \left(\frac{Re}{2}\right)^2 = \frac{Re^2}{4} \]
Therefore, we can rewrite the force on the planet as:
\[ F_g \text{ (on planet)} = \frac{G \cdot (2Me) \cdot m}{\frac{Re^2}{4}} = \frac{G \cdot (2Me) \cdot m \cdot 4}{Re^2} \]
This simplifies to:
\[ F_g \text{ (on planet)} = \frac{8G \cdot Me \cdot m}{Re^2} \]

6. **Relate it to the Force on Earth:**
We know that \( F = \frac{G \cdot Me \cdot m}{Re^2} \). Thus, we can substitute this into our equation:
\[ F_g \text{ (on planet)} = 8 \cdot F \]

### Final Answer:
The force of attraction on the same object on the surface of that planet is \( 8F \).

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