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)

15√v
the fraction with numerator 1 and denominator square root of 5 v

15v
1 fifth v

5–√v
square root of 5 v

5v

5v

The correct answer is 5v.

To answer this question, we can use Kepler's Third Law of Planetary Motion. According to Kepler's Third Law, the square of the orbital period of a satellite is proportional to the cube of its average orbital radius. In this case, since we are given the ratio of the orbital radii, we can use the square root of this ratio to find the ratio of the orbital periods.

The orbital period, T, is directly related to the velocity, v, by the formula:

v = (2πr) / T

Let's denote the initial orbital radius as r0 and the initial velocity as v0. The ratio of the initial and final orbital radii can be written as:

r_final / r_initial = 5r / r = 5

Taking the square root of both sides, we get:

√(r_final / r_initial) = √5

Now, let's denote the final velocity as v_final. According to Kepler's Third Law, the ratio of the orbital periods is equal to the ratio of the average velocities. Therefore, we have:

T_final / T_initial = v_final / v_initial

Since T_final is directly proportional to r_final (and T_initial is directly proportional to r_initial), we can substitute the ratios of the orbital radii into the equation:

√5 = v_final / v_initial

Next, solve for v_final:

v_final = √5 * v_initial

In this case, the initial velocity is v, so the final velocity is:

v_final = √5 * v

Therefore, the correct answer is:

5–√v