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

Question:

A satellite is in a circular orbit around an unknown planet. The satellite has a speed of 1.80*10{eq}^4 {/eq} m/s, and the radius of the orbit is 5.20 106 m. A second satellite also has a circular orbit around this same planet. The orbit of this second satellite has a radius of 8.70*10{eq}^6 {/eq} m. What is the orbital speed of the second satellite?

Orbital speed:

The orbital speed of the satellite is defined as the speed at which a satellite revolves about another body. Orbital speed is also known as the minimum velocity required to place a satellite in the orbit.

Answer and Explanation:


Given Data

  • The speed of the first satellite is: {eq}{v_1} = 1.80 \times {10^4}\;{\rm{m/s}} {/eq}.
  • The radius of the first orbit is: {eq}{r_1} = 5.20 \times {10^6}\;{\rm{m}} {/eq}.
  • The speed of the second satellite is: {eq}{r_2} = 8.70 \times {10^6}\;{\rm{m}} {/eq}.


The expression for the radius of the second satellite is,

{eq}\begin{align*} \dfrac{{{v_2}}}{{{v_1}}} &= \sqrt {\dfrac{{{r_1}}}{{{r_2}}}} \\ {v_2} &= {v_1}\sqrt {\dfrac{{{r_1}}}{{{r_2}}}} \end{align*} {/eq}


Substitute the given values.

{eq}\begin{align*} {v_2} &= \left( {1.80 \times {{10}^4}\;{\rm{m/s}}} \right)\sqrt {\dfrac{{\left( {5.20 \times {{10}^6}\;{\rm{m}}} \right)}}{{\left( {8.70 \times {{10}^6}\;{\rm{m}}} \right)}}} \\ &= \left( {1.80 \times {{10}^4}\;{\rm{m/s}}} \right)\left( {0.773} \right)\\ &= 1.3914 \times {10^4}\;{\rm{m/s}} \end{align*} {/eq}


Thus, the orbital speed of the satellite is {eq}1.3914 \times {10^4}\;{\rm{m/s}} {/eq}.


Learn more about this topic:

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Elliptical Orbits: Periods & Speeds

from Physics: High School

Chapter 7 / Lesson 11
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