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You are standing on the surface of a planet that has spherical symmetry and a radius of 5.00×10⁶m. The gravitational potential energy u of the system composed of you and the planet is -1.20×10⁹J if we choose u to be zero when you are very far from the planet. Part a) What is the magnitude of the gravitational force that the planet exerts on you when you are standing on its surface?

User PooSH
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Final answer:

The magnitude of the gravitational force that the planet exerts on you when you are standing on its surface is approximately 2.39 x 10^14 N.

Step-by-step explanation:

To calculate the magnitude of the gravitational force that the planet exerts on you when you are standing on its surface, we can use Newton's universal law of gravitation. The formula is F = (G * m1 * m2) / r^2, where F is the force of gravity, G is the gravitational constant (approximately 6.67 x 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the two objects (you and the planet), and r is the distance between the centers of the two objects (equal to the planet's radius in this case).

In this scenario, since you are standing on the surface of the planet, your mass is negligible compared to the planet's mass. Therefore, you can assume that m1 (your mass) is very small and only consider the mass of the planet (m2). Substituting the values into the formula, we get:

F = (6.67 x 10^-11 Nm^2/kg^2) * (m1 * 5.97 x 10^24 kg) / (5.00 x 10^6 m)^2

F = 5.97 x 10^24 kg * (6.67 x 10^-11 Nm^2/kg^2) / (5.00 x 10^6 m)^2

F = 5.97 x 10^24 kg * (6.67 x 10^-11 Nm^2/kg^2) / 2.50 x 10^13 m^2

F = 2.39 x 10^14 N

Therefore, the magnitude of the gravitational force that the planet exerts on you when you are standing on its surface is approximately 2.39 x 10^14 N.

User Faraday
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