Find the orbital speed of a satellite in a circular orbit 3.70 107 m above the surface of the...

Question:

Find the orbital speed of a satellite in a circular orbit 3.70 107 m above the surface of the Earth. v = 3030 m/s

Orbital Speed

Newton's laws govern the motions of objects on earth which extends to the motion of planets, moons and other satellites. The mathematics used for circular motion is the same mathematics that describes a satellite's motion. Orbital speed is can be calculated using:

$$v = \sqrt \frac {G M_c}{R} $$ where {eq}G = 6.673 \times 10^{-11} N m^2 / kg^2 {/eq}, {eq}M_c {/eq} is the mass of the center object about which the satellite orbits and {eq}R {/eq} is the radius of orbit for a given satellite.

Answer and Explanation:

Given:

{eq}G = 6.673 \times 10^\text{-11} N m^2 / kg^2 \\ R_s = 3.70 \times 10^\text{7} m\\ R_E = 6.37 \times 10^\text{6} m \\ M_E = 5.98\times 10^\text{24} kg \\ {/eq}

Solution:

{eq}v = \sqrt \frac {G M_E}{R_E + R_s} = \sqrt \frac {(6.673 \times 10^\text{-11} N m^2 / kg^2 ) (5.98\times 10^\text{24} kg )}{6.37 \times 10^\text{6} m + 3.70 \times 10^\text{7} m} = 3033 m/s {/eq}


Learn more about this topic:

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How Orbits Are Influenced by Gravity & Energy

from Basics of Astronomy

Chapter 25 / Lesson 6
15K

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