A body on the surface of the Earth weighs 400 lb.The radius of the Earth is about 4,000 miles.If this same body were placed on a 4,000-mile-high tower,its weight would be

Question

Quizplus · Accepted Answer

The weight of an object on the surface of the Earth is given by the formula W =mg, where m is the mass of the object and g is the acceleration due to gravity. On the surface of the Earth, g is approximately 32 feet per second squared. Therefore, the weight of the body on the surface of the Earth is:

W = mg
W = (mass of body)(acceleration due to gravity)
W = (400 lb)(32 ft/s^2)
W ≈ 12,800 lb

If the same body were placed on a 4,000-mile-high tower, it would be 4,000 miles farther from the center of the Earth. The gravitational force on the body would be weaker, and therefore its weight would be less. The acceleration due to gravity at a distance R from the center of the Earth is given by the formula g = G (M/R^2), where G is the gravitational constant, M is the mass of the Earth, and R is the distance from the center of the Earth.

At the surface of the Earth, R = 4,000 miles, so we have:

g = G (M/R^2)
g = (6.67 × 10^-11 m^3/kg s^2)(5.97 × 10^24 kg)/(6,371,000 m)^2
g ≈ 9.81 m/s^2 (approximately 32 ft/s^2)

At a distance of 8,000 miles (4,000 miles above the surface), we have:

g = G (M/R^2)
g = (6.67 × 10^-11 m^3/kg s^2)(5.97 × 10^24 kg)/(8,000,000 m)^2
g ≈ 4.91 m/s^2 (approximately 16 ft/s^2)

Therefore, the weight of the body on the tower would be:

W = mg
W = (mass of body)(acceleration due to gravity)
W = (400 lb)(16 ft/s^2)
W ≈ 6,400 lb

Thus, the weight of the body on the 4,000-mile-high tower would be 100 lbs., approximately.

A Body on the Surface of the Earth Weighs 400