Final answer:
The weight of an object decreases with the square of the distance from the center of the mass it is orbiting. At 3R, the distance from the Earth's center, the gravitational force would be 1/9th of what it is on Earth's surface. Hence, an object that weighs 180 N on Earth would weigh 20 N at a distance of 3R from Earth's center.
Step-by-step explanation:
To determine the weight of an object at a distance of 3R from the center of Earth, we need to understand how gravitational force decreases with distance. The gravitational force and, therefore, gravitational acceleration (g) at a distance r from a mass M is given by Newton's law of universal gravitation:
F = G(Mm/r²)
where G is the gravitational constant, M is the mass of Earth, m is the mass of the object, and r is the distance to the center of Earth. The weight (W) of an object is the product of its mass m and the gravitational acceleration g, so:
W = mg
The gravitational acceleration (g) on Earth's surface is 9.80 m/s². We can use the formula:
g' = GM/(3R)²
to find the gravitational acceleration at a distance of 3R. Since the mass of the object cancels out, we get:
g' = g/R²
And we know the object's weight on Earth's surface (W = 180 N), so:
W' = (W/R²) x 1/9
Substituting the values, we have:
W' = 180 N / 9 = 20 N
Therefore, at a distance of 3R from the center of the Earth, the object would weigh 20 N.