A new planet is discovered that has twice the Earth’s mass and twice the Earth’s radius. On the surface of...

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A new planet is discovered that has twice the Earth’s mass and twice the Earth’s radius. On the surface of this new planet, a person who weighs 500 N on Earth would experience a gravitational force of

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To find the gravitational force experienced by a person on the surface of the new planet, we can use the formula for gravitational force: $$ F = \frac{G \cdot m_1 \cdot m_2}{r^2} $$ where: - $ F $ is the gravitational force, - $ G $ is the gravitational constant, approximately $ 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 $, - $ m_1 $ is the mass of the planet, - $ m_2 $ is the mass of the person (which we can derive from their weight), - $ r $ is the radius of the planet. 1. **Calculating the mass of the person:** On Earth, the relationship between weight (W) and mass (m) is given by: $$ W = m \cdot g $$ where: - $ g $ on Earth is approximately $ 9.81 \, \text{m/s}^2 $. Rearranging the formula gives us: $$ m = \frac{W}{g} = \frac{500 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 51.03 \, \text{kg} $$ 2. **Defining the properties of the new planet:** - Mass of the new planet ($ m_{planet} $): Since the new planet has twice the mass of Earth, if the mass of Earth ($ m_{Earth} $) is approximately $ 5.972 \times 10^{24} \, \text{kg} $: $$ m_{planet} = 2 \cdot m_{Earth} = 2 \cdot 5.972 \times 10^{24} \approx 1.1944 \times 10^{25} \, \text{kg} $$ - Radius of the new planet ($ r_{planet} $): The new planet has twice the radius of Earth, where the radius of Earth ($ r_{Earth} $) is approximately $ 6.371 \times 10^6 \, \text{m} $: $$ r_{planet} = 2 \cdot r_{Earth} = 2 \cdot 6.371 \times 10^6 \approx 1.2742 \times 10^7 \, \text{m} $$ 3. **Calculating the gravitational force on the new planet:** The acceleration due to gravity on the new planet ($ g_{planet} $) can be calculated using: $$ g_{planet} = \frac{G \cdot m_{planet}}{r_{planet}^2} $$ Substituting in values: $$ g_{planet} = \frac{(6.674 \times 10^{-11}) \cdot (1.1944 \times 10^{25})}{(1.2742 \times 10^7)^2} $$ Calculating $ (1.2742 \times 10^7)^2 $: $$ (1.2742 \times 10^7)^2 \approx 1.6225 \times 10^{14} \, \text{m}^2 $$ Now, substituting this into the gravitational equation give...

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A new planet is discovered that has twice the Earth’s mass and twice the Earth’s radius. On the surface of...

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