Posted by Jessy on Monday, January 15, 2007 at 3:10pm.
A orbiting satellite stays over a certain spot on the equator of (rotating) Venus. What is the altitude of the orbit (called a "synchronous orbit")?
I found the radius and mass of Venus on the google, but I don't know what I should plug in the "T" if I use the Kepler's 3rd law.
Or are there any other ways to solve?
Thank you
T is the time needed for venus to rotate about it axis which is 243.02 earth days.
So how to convert this to seconds?
just simply times 24 hours times 3600 seconds?
The answer is that such an orbit does not exist. If you solve for the radius you find that it is about 1.5 million kilometers, but at that distance from Venus, the gravitational force exerted by the sun is about 40 times stronger than the gravitational force exerted by Venus.
Yes, convert to seconds and plug that in the formula:
T = 2 pi r^(3/2)/sqrt[MG]
Solve for r.
Actually the gravitaional force exerted of the sun on the satellite is not relevant. We need to evaluate the tidal force, which is the gravitational acceleration in the Sun's gravitational field at Venus minus that at the satellite's position times the mass of the satellite.
Venus, just like the satellite is freely falling toward the sun. Relative to an observer on Venus there seems to exists a socalled fictitious tidal force that is perturbing the orbit of the satellite.
I find that this tidal force exerted by the Sun is twice the gravitational force exerted by Venus if the satellite is is in the direction of the sun or opposite to it aas seen from Venus.
So, one can conclude that the orbit doesn't exist.

ybacluqf gqwv  ybacluqf gqwv, Sunday, December 28, 2008 at 8:35am
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