Question
A satellite orbiting Earth at an orbital radius r has a velocity v. What would the velocity be if the satellite is moved to an orbital radius of 6r?(1 point)
Responses
16v
1
6
𝑣
1 sixth v
6v
6
𝑣
6 v
16√v
1
6
𝑣
the fraction with numerator 1 and denominator square root of 6 v
6–√v
6
𝑣
square root of 6 v
Responses
16v
1
6
𝑣
1 sixth v
6v
6
𝑣
6 v
16√v
1
6
𝑣
the fraction with numerator 1 and denominator square root of 6 v
6–√v
6
𝑣
square root of 6 v
Answers
GPT-4o mini
To find the new velocity of a satellite when it is moved to an orbital radius of \(6r\) from its original radius \(r\), we can use the formula for orbital velocity:
\[
v = \sqrt{\frac{GM}{r}}
\]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the orbital radius.
If the satellite is moved to an orbital radius of \(6r\), the new velocity \(v'\) can be calculated as follows:
\[
v' = \sqrt{\frac{GM}{6r}} = \sqrt{\frac{1}{6}} \sqrt{\frac{GM}{r}} = \sqrt{\frac{1}{6}} v
\]
Thus, the new velocity \(v'\) is:
\[
v' = \frac{v}{\sqrt{6}}
\]
So the correct response is:
\[
\text{the fraction with numerator 1 and denominator } \sqrt{6} v
\]
\[
v = \sqrt{\frac{GM}{r}}
\]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the orbital radius.
If the satellite is moved to an orbital radius of \(6r\), the new velocity \(v'\) can be calculated as follows:
\[
v' = \sqrt{\frac{GM}{6r}} = \sqrt{\frac{1}{6}} \sqrt{\frac{GM}{r}} = \sqrt{\frac{1}{6}} v
\]
Thus, the new velocity \(v'\) is:
\[
v' = \frac{v}{\sqrt{6}}
\]
So the correct response is:
\[
\text{the fraction with numerator 1 and denominator } \sqrt{6} v
\]