E Native American mascots - hor x SI Course Modules: AST 111: Intro x A Ch 21: Venus and Mars - AST 1 x © Squaring both sides and solvin x| + A webassign.net/web/Student/Assignment-Responses/submit?pos=16&dep=24621113&tags=autosave#question4793215_16 Tutorial The Magellan orbiter orbits Venus with a period of 3.26 hours. How far (in km) above the surface of the planet is it? (The mass of Venus is 4.87 x 1024 kg, and the radius of Venus is 6.05 x 103 km.) Part 1 of 3 The period of the orbiter's orbit can give us the speed at which the orbiter orbits the planet. We imagine the orbiter tracing a circle around the planet at a certain height, the speed is 2ar V = P Part 2 of 3 Next, we combine this with the circular velocity equation to determine the height above the planet's surface. GM V = 2ar GM Squaring both sides and solving for r gives the following equation. What is the exponent for r? GM Part 3 of 3 Congratulations! You just derived a version of Kepler's Third Law for Venus! Using the mass of Venus in kilograms and converting the 3.26 hours to seconds, calculate the distance from the center of the planet. GMg r = km And then determine the distance from the surface. r = ry+'s 's km

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The Magellan orbiter orbits Venus with a period of 3.26 hours. How far (in km) above the surface of the planet is it? (The mass of Venus is 4.87 ✕ 1024 kg, and the radius of Venus is 6.05 ✕ 103 km.)

E Native American mascots - hor x SI Course Modules: AST 111: Intro x
A Ch 21: Venus and Mars - AST 1 x © Squaring both sides and solvin x| +
A webassign.net/web/Student/Assignment-Responses/submit?pos=16&dep=24621113&tags=autosave#question4793215_16
Tutorial
The Magellan orbiter orbits Venus with a period of 3.26 hours. How far (in km) above the surface of the planet
is it? (The mass of Venus is 4.87 x 1024 kg, and the radius of Venus is 6.05 x 103 km.)
Part 1 of 3
The period of the orbiter's orbit can give us the speed at which the orbiter orbits the planet. We imagine the
orbiter tracing a circle around the planet at a certain height, the speed is
2ar
V =
P
Part 2 of 3
Next, we combine this with the circular velocity equation to determine the height above the planet's surface.
GM
V =
2ar
GM
Squaring both sides and solving for r gives the following equation. What is the exponent for r?
GM
Part 3 of 3
Congratulations! You just derived a version of Kepler's Third Law for Venus!
Using the mass of Venus in kilograms and converting the 3.26 hours to seconds, calculate the distance from
the center of the planet.
GMg
r =
km
And then determine the distance from the surface.
r = ry+'s
's
km
Transcribed Image Text:E Native American mascots - hor x SI Course Modules: AST 111: Intro x A Ch 21: Venus and Mars - AST 1 x © Squaring both sides and solvin x| + A webassign.net/web/Student/Assignment-Responses/submit?pos=16&dep=24621113&tags=autosave#question4793215_16 Tutorial The Magellan orbiter orbits Venus with a period of 3.26 hours. How far (in km) above the surface of the planet is it? (The mass of Venus is 4.87 x 1024 kg, and the radius of Venus is 6.05 x 103 km.) Part 1 of 3 The period of the orbiter's orbit can give us the speed at which the orbiter orbits the planet. We imagine the orbiter tracing a circle around the planet at a certain height, the speed is 2ar V = P Part 2 of 3 Next, we combine this with the circular velocity equation to determine the height above the planet's surface. GM V = 2ar GM Squaring both sides and solving for r gives the following equation. What is the exponent for r? GM Part 3 of 3 Congratulations! You just derived a version of Kepler's Third Law for Venus! Using the mass of Venus in kilograms and converting the 3.26 hours to seconds, calculate the distance from the center of the planet. GMg r = km And then determine the distance from the surface. r = ry+'s 's km
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