# A satellite is in a circular orbit around an unknown planet. The satellite has a speed of 1.70 x...

## Question:

A satellite is in a circular orbit around an unknown planet. The satellite has a speed of {eq}1.70 \times 10^4 {/eq} m/s, and the radius of the orbit is {eq}5.80 \times 10^6 {/eq} m. A second satellite also has a circular orbit around this same planet. The orbit of this second satellite has a radius of {eq}8.50 \times 10^6 {/eq} m. What is the orbital speed of the second satellite?

## Orbital speed in a circular orbit

If a body moves in a circular orbit with radius {eq}r {/eq} around a more massive body with mass {eq}M {/eq}, its orbital speed is

{eq}v = \sqrt{\dfrac{G M}{r}}, {/eq}

where {eq}G = 6.67 \times 10^{-11} \ \text{N} \ \text{m}^2/\text{kg}^2 {/eq} is the gravitation constant.

## Answer and Explanation:

Notice that we can write

{eq}v = \sqrt{\dfrac{G M}{r}} \quad \Rightarrow \quad v \sqrt{r} = \sqrt{GM}. {/eq}

Thus, for a planet with mass {eq}M {/eq}, the orbital speed {eq}v {/eq} of a satellite in a circular orbit with radius {eq}r {/eq} satisfies {eq}v \sqrt{r} = \text{constant} {/eq}. This means that at radii {eq}r_1 {/eq} and {eq}r_2 {/eq}, the orbital speeds of the satellites, {eq}v_1 {/eq} and {eq}v_2 {/eq} respectively, satisfy

{eq}v_1 \sqrt{r_1} = v_2 \sqrt{r_2} \quad \Rightarrow \quad v_2 = v_1 \sqrt{\dfrac{r_1}{r_2}}. {/eq}

Thus, with {eq}v_1 = 1.70 \times 10^4 \ \text{m}/\text{s} {/eq}, {eq}r_1 = 5.80 \times 10^6 \ \text{m} {/eq}, then the orbital speed {eq}v_2 {/eq} at {eq}r_2 = 8.50 \times 10^6 \ \text{m} {/eq} is

{eq}v_2 = (1.70 \times 10^4 \ \text{m}/\text{s}) \sqrt{\dfrac{5.80 \times 10^6 \ \text{m}}{8.50 \times 10^6 \ \text{m}}} = 1.40 \times 10^4 \ \text{m}/\text{s}. {/eq}

Therefore, the orbital speed of the satellite at {eq}8.50 \times 10^6 \ \text{m} {/eq} is {eq}1.40 \times 10^4 \ \text{m}/\text{s} {/eq}.

#### Learn more about this topic:

Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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