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What is the gravitational force exerted on a 74.2 kg person standing on Earth's surface, a distance of 6.39 x 10^6 m from the center of the planet?

a) 727.2 N
b) 9.80 N
c) 74.2 N
d) 4.65 x 10^9 N

User Jonnysamps
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1 Answer

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Final answer:

The gravitational force exerted on a 74.2 kg person standing on Earth's surface is 727.2 N.

Step-by-step explanation:

The gravitational force exerted on a 74.2 kg person standing on Earth's surface can be calculated using Newton's universal law of gravitation. The formula is given as:

F = (G * M * m) / (r^2)

Where F is the gravitational force, G is the gravitational constant (6.674 × 10^-11 N·m² kg^2), M is the mass of the planet (5.97 × 10^24 kg), m is the mass of the person (74.2 kg), and r is the distance from the center of the planet (6.39 × 10^6 m).

Plugging in the values:

F = (6.674 × 10^-11 N·m² kg^2 * 5.97 × 10^24 kg * 74.2 kg) / (6.39 × 10^6 m)^2

After calculating, the answer is 727.2 N, which corresponds to option a).

User DJeePe
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