When an astronaut is on a spacewalk and is moving away from the space shuttle, how does the force of gravity between the two objects change? (1 point)

Question

When an astronaut is on a spacewalk and is moving away from the space shuttle, how does the force of gravity between the two objects change? (1 point)
O it goes down for the astronaut and goes up for the space shuttle
• it goes down
• it goes up for the astronaut and goes down for the space shuttle
• it goes up

GPT-4o mini · Answer

The correct answer is: **it goes down**. The force of gravity between two objects, such as an astronaut and a space shuttle, is governed by Newton's law of universal gravitation, which states that the gravitational force depends on the mass of the objects and the distance between them. When the astronaut moves away from the space shuttle, the distance between the two increases, and since gravitational force decreases with increasing distance, the gravitational force between the astronaut and the space shuttle decreases.

AYD3N · Answer

How does the gravitational force between Earth and the Moon change when the distance between the two objects doubles but mass doesn't change? (1 point)
The gravitational force between Earth and the Moon would goes up by a proportional amount.
The gravitational force between Earth and the Moon would stay the
same.
The gravitational force between Earth and the Moon would completely disappear.
• The gravitational force between Earth and the Moon would go down
by a proportional amount.

GPT-4o mini · Answer

The correct answer is: **The gravitational force between Earth and the Moon would go down by a proportional amount.** According to Newton's law of universal gravitation, the gravitational force $ F $ between two masses is given by the equation: $$ F = \frac{G \cdot m_1 \cdot m_2}{r^2} $$ where $ G $ is the gravitational constant, $ m_1 $ and $ m_2 $ are the masses of the two objects, and $ r $ is the distance between their centers. If the distance $ r $ doubles, the gravitational force becomes: $$ F' = \frac{G \cdot m_1 \cdot m_2}{(2r)^2} = \frac{G \cdot m_1 \cdot m_2}{4r^2} = \frac{1}{4}F $$ Thus, the gravitational force decreases to a quarter of its original value, which means it goes down by a proportional amount.