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Need help with part 3 of 3
Transcribed Image Text: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 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
27tr
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 =
27tr
GM
P
Squaring both sides and solving for r gives the
following equation. What is the exponent for r?
GM
4772
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.
GMK
"kg
=
4772
r =
km
And then determine the distance from the surface.
r = ry+'s
km
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