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A satellite orbiting Earth at an orbital radius r has a velocity v. Which represents the velocity if the satellite is moved to...Question
A satellite orbiting Earth at an orbital radius r has a velocity v. Which represents the velocity if the satellite is moved to an orbital radius of 5r?(1 point)
Responses
1/5√v
the fraction with numerator 1 and denominator square root of 5 v
1/5v
1 fifth v
5–√v
square root of 5 v
5v
Responses
1/5√v
the fraction with numerator 1 and denominator square root of 5 v
1/5v
1 fifth v
5–√v
square root of 5 v
5v
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Answered by
GPT-4o mini
To determine the velocity \( v' \) of a satellite moved from an orbital radius \( r \) to an orbital radius of \( 5r \), we can use the formula for the orbital velocity of a satellite, which is given by:
\[
v = \sqrt{\frac{GM}{r}}
\]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.
For the new orbital radius \( 5r \):
\[
v' = \sqrt{\frac{GM}{5r}} = \sqrt{\frac{1}{5}} \sqrt{\frac{GM}{r}} = \frac{1}{\sqrt{5}} v
\]
Thus, the velocity of the satellite at the orbit of radius \( 5r \) is:
\[
v' = \frac{1}{\sqrt{5}} v
\]
So, the correct response is:
**1/5√v**
\[
v = \sqrt{\frac{GM}{r}}
\]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.
For the new orbital radius \( 5r \):
\[
v' = \sqrt{\frac{GM}{5r}} = \sqrt{\frac{1}{5}} \sqrt{\frac{GM}{r}} = \frac{1}{\sqrt{5}} v
\]
Thus, the velocity of the satellite at the orbit of radius \( 5r \) is:
\[
v' = \frac{1}{\sqrt{5}} v
\]
So, the correct response is:
**1/5√v**
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