Question
I satellite warping Earth at an open radius r has a velocity v which represents the velocity of the satellite is moved to an orbital radius of 5r answer 1 1/5 v square root 5v5b
Answers
GPT-4o mini
To analyze the problem of a satellite changing its orbital radius and the subsequent effects on its velocity, we can use the principles of orbital mechanics.
In a circular orbit, the gravitational force provides the necessary centripetal force for the satellite to remain in orbit. The gravitational force can be described by the formula:
\[
F = \frac{G M m}{r^2}
\]
where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the radius of the orbit.
The centripetal force required to keep the satellite in a circular orbit is given by:
\[
F = \frac{m v^2}{r}
\]
where:
- \( v \) is the orbital velocity of the satellite.
Setting these two forces equal to each other for a satellite in a circular orbit, we get:
\[
\frac{G M m}{r^2} = \frac{m v^2}{r}
\]
After canceling \( m \) from both sides (assuming \( m \neq 0 \)), we can rearrange the equation to find \( v \):
\[
v^2 = \frac{G M}{r}
\]
Now, considering the scenario where the satellite initially is in an orbit with radius \( r \) and velocity \( v \), and is moved to an orbit with radius \( 5r \):
For the new orbit at radius \( 5r \), we can apply the same formula:
\[
v' = \sqrt{\frac{G M}{5r}}
\]
Since \( v^2 = \frac{G M}{r} \), we can substitute this into the equation for \( v' \):
\[
v' = \sqrt{\frac{1}{5}} \sqrt{\frac{G M}{r}} = \sqrt{\frac{1}{5}} v = \frac{v}{\sqrt{5}}
\]
Thus, the velocity of the satellite when moved to an orbital radius of \( 5r \) is:
\[
v' = \frac{v}{\sqrt{5}}
\]
So, from the options you provided, the answer is:
\[
\frac{1}{\sqrt{5}} v
\]
If you have specific options that you would like me to analyze further or if there are further details, please let me know!
In a circular orbit, the gravitational force provides the necessary centripetal force for the satellite to remain in orbit. The gravitational force can be described by the formula:
\[
F = \frac{G M m}{r^2}
\]
where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the radius of the orbit.
The centripetal force required to keep the satellite in a circular orbit is given by:
\[
F = \frac{m v^2}{r}
\]
where:
- \( v \) is the orbital velocity of the satellite.
Setting these two forces equal to each other for a satellite in a circular orbit, we get:
\[
\frac{G M m}{r^2} = \frac{m v^2}{r}
\]
After canceling \( m \) from both sides (assuming \( m \neq 0 \)), we can rearrange the equation to find \( v \):
\[
v^2 = \frac{G M}{r}
\]
Now, considering the scenario where the satellite initially is in an orbit with radius \( r \) and velocity \( v \), and is moved to an orbit with radius \( 5r \):
For the new orbit at radius \( 5r \), we can apply the same formula:
\[
v' = \sqrt{\frac{G M}{5r}}
\]
Since \( v^2 = \frac{G M}{r} \), we can substitute this into the equation for \( v' \):
\[
v' = \sqrt{\frac{1}{5}} \sqrt{\frac{G M}{r}} = \sqrt{\frac{1}{5}} v = \frac{v}{\sqrt{5}}
\]
Thus, the velocity of the satellite when moved to an orbital radius of \( 5r \) is:
\[
v' = \frac{v}{\sqrt{5}}
\]
So, from the options you provided, the answer is:
\[
\frac{1}{\sqrt{5}} v
\]
If you have specific options that you would like me to analyze further or if there are further details, please let me know!