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What is the gravitational force acting on a 70.0 kg object standing on the Earth’s surface?

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User Mosby
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Answer:

What is the gravitational force acting on a 70.0 kg object standing on the Earth’s surface?

Explanation:

The gravitational force acting on a 70.0 kg object standing on the Earth’s surface can be calculated using the formula:

F = G * (m1 * m2) / r^2

Where:

F = gravitational force

G = gravitational constant (6.6743 x 10^-11 N*m^2/kg^2)

m1 = mass of the first object (in this case, the mass of the Earth)

m2 = mass of the second object (in this case, the mass of the object, 70.0 kg)

r = distance between the centers of the two objects (in this case, the radius of the Earth, which is approximately 6,371 km)

Using the values above, we can calculate the gravitational force as follows:

F = (6.6743 x 10^-11 N*m^2/kg^2) * (5.97 x 10^24 kg * 70.0 kg) / (6,371,000 m)^2

F = 686.6 N

Therefore, the gravitational force acting on a 70.0 kg object standing on the Earth’s surface is approximately 686.6 N.

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User Victor Grazi
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