Question

The weight of an object varies inversely with the second power of the distance from the center of Earth. The radius of Earth is approximately 4000 miles. If a person weighs 160 pounds on Earth's surface, what would this individual weigh 8000 miles above the surface of Earth?

Answer 1

Weight = 17.8 pounds

Step-by-step explanation:

Given that the weight W of a person is inversely proportional to the distance of that person from the center of the earth, then W ∝1/d²

and W2/W1 = d1²/d2²

So W2 = W1× d1²/d2²

Given W1 = 160lb, d1 = 4000miles, d2 = (8000 + 4000) = 12000miles. This is because this distance is always measured from the center of the earth. At the earth's surface the distance from the earth's center is 4000miles so at a distance of 8000miles from the earth's surface the distance from the center would now be the addition of both distances.

So W2 = 160× 4000²/12000² = 17.8lb.

This law in operation is known as the inverse square law.

The weight is actually the force of gravity between the earth and the individual. It is an attractive force that decreases with increasing distance.

W = GMem/d²

Where G = gravitational constant = 6.67×10-¹¹ Nm²/kg²

Me = mass of the earth

m = mass of individual

d = distance from the center of the earth.

Answer 2

Answer: 17.8 pounds.

Step-by-step explanation:

The radius of the Earth is 4000 mi.

now, a person right in the surface of the earth is located at a distance 4000 miles of the center, and we have that the weight will be something like:

W = 160 pounds = A/(4000)^2 pounds

where A is a constant that we do not know, but we can find the value:

A = 160*(4000)^2

Now, at a distance of 8000 miles to the surface we have:

W = A/(4000 + 8000)^2 = 160*(4000)^2/(12000)^2 = 17.8 pounds.