TutorialThe Magellan orbiter orbits Venus with a period of 3.26hours. How far (in km) above the surface of the planetis it? (The mass of Venus is 4.87 × 1024 kg, and theradius of Venus is 6.05 × 103 km.)Part 1 of 3The period of the orbiter's orbit can give us the speedat which the orbiter orbits the planet. We imagine theorbiter tracing a circle around the planet at a certainheight, the speed is27trV =Part 2 of 3Next, we combine this with the circular velocityequation to determine the height above the planet'ssurface.GMV =27trGMSquaring both sides and solving for r gives thefollowing equation. What is the exponent for r?U. GM472

The equation for the radius of the orbit is given by, r3=GMP24π2 Substituting the values in the equation, r3=6.67×10-11 N·m2/kg24.87×1024 kg3.26×3600 s24π2r=6.67×10-11 N·m2/kg24.87×1024 kg3.26×3600 s24π213=10.42×106 m=10.42×103 kmThe radius of the orbit is the distance from the center of the planet to the center of the orbiter. The height of the center of the orbiter from the surface of the planet is given by, h=r-R Here, R is the radius of the planet. Substituting, h=10.42×103 km-6.05×103 km=4.4×103 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.)

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.)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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