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

A) 1/5√v
the fraction with numerator 1 and denominator square root of 5 v

B) 1/5v
1 fifth v

C) √5v
square root of 5 v

D) 5v

Answer 1

To determine the velocity of a satellite in orbit as its orbital radius changes, we can use the formula for the orbital velocity of a satellite in a circular orbit, which is given by:

\[
v = \sqrt{\frac{GM}{r}}
\]

where:
- \( v \) is the orbital velocity,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( r \) is the orbital radius.

If we denote the initial orbital radius as \( r \) and the corresponding velocity as \( v \), then:

\[
v = \sqrt{\frac{GM}{r}}
\]

Now, if the satellite is moved to an orbital radius of \( 5r \), we can find the new velocity \( v' \):

\[
v' = \sqrt{\frac{GM}{5r}} = \sqrt{\frac{1}{5}} \sqrt{\frac{GM}{r}} = \frac{1}{\sqrt{5}} v
\]

Thus, the velocity at the new radius (5r) is:

\[
v' = \frac{1}{\sqrt{5}} v
\]

This means that the correct answer is:

**A) \( \frac{1}{5\sqrt{v}} \)**

However, because A) as written seems incorrect, the expression should actually indicate that the velocity is proportional to \( \frac{1}{\sqrt{5}} v \), so none of the options in the list perfectly matches this conclusion.

If the intended meaning of option A may have a contextual error on representation, the final expression shows that the new orbital velocity is indeed \( \frac{1}{\sqrt{5}}v \), aligning closest with Option A as needing clarification. Thus, please check the representation carefully against the options provided.