Question
Two satellites of equal mass are in circular orbit around a planet. Satellite b is twice as far away. Which feels a stronger gravitational force? (1 point)
Satellite A
Satellie B
Both are the same
Neither feels a gravitational force
Satellite A
Satellie B
Both are the same
Neither feels a gravitational force
Answers
GPT-4o mini
The gravitational force between two objects is given by Newton's law of universal gravitation, which states that the force \( F \) is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is:
\[
F = G \frac{m_1 \cdot m_2}{r^2}
\]
Where:
- \( G \) is the gravitational constant,
- \( m_1 \) and \( m_2 \) are the masses of the two objects,
- \( r \) is the distance between their centers.
In this case, both satellites have the same mass, say \( m \). Let's denote the distance of Satellite A from the planet as \( r \), and for Satellite B, it is \( 2r \).
Now, the gravitational force on Satellite A:
\[
F_A = G \frac{M \cdot m}{r^2}
\]
And the gravitational force on Satellite B:
\[
F_B = G \frac{M \cdot m}{(2r)^2} = G \frac{M \cdot m}{4r^2}
\]
Comparing the two forces:
- \( F_A \) is proportional to \( \frac{1}{r^2} \).
- \( F_B \) is proportional to \( \frac{1}{4r^2} \), which is one-fourth of \( F_A \).
Thus, Satellite A experiences a stronger gravitational force because it is closer to the planet.
**Answer: Satellite A**.
\[
F = G \frac{m_1 \cdot m_2}{r^2}
\]
Where:
- \( G \) is the gravitational constant,
- \( m_1 \) and \( m_2 \) are the masses of the two objects,
- \( r \) is the distance between their centers.
In this case, both satellites have the same mass, say \( m \). Let's denote the distance of Satellite A from the planet as \( r \), and for Satellite B, it is \( 2r \).
Now, the gravitational force on Satellite A:
\[
F_A = G \frac{M \cdot m}{r^2}
\]
And the gravitational force on Satellite B:
\[
F_B = G \frac{M \cdot m}{(2r)^2} = G \frac{M \cdot m}{4r^2}
\]
Comparing the two forces:
- \( F_A \) is proportional to \( \frac{1}{r^2} \).
- \( F_B \) is proportional to \( \frac{1}{4r^2} \), which is one-fourth of \( F_A \).
Thus, Satellite A experiences a stronger gravitational force because it is closer to the planet.
**Answer: Satellite A**.