A satellite GeoSAT is in a circular geostationary orbit of radius RG, above a point P
on the equator. Another satellite ComSAT is in a lower circular orbit of radius 0.81RG.
At 7 P.M on January 1, ComSAT is sighted directly above P. On which DATE can ComSAT be sighted directly above P between 7 P.M. and 8 P.M.?

Question

A satellite GeoSAT is in  a  circular  geostationary  orbit  of  radius RG, above a point P
on the equator. Another satellite ComSAT is in a lower circular orbit of radius 0.81RG.
At 7 P.M on January 1, ComSAT is sighted directly above P. On which DATE can ComSAT be sighted directly above P between 7 P.M. and 8 P.M.?

jamesj · Answer

By one of Kepler's laws, the period T of an orbit and the semi-major axis of that orbit R (or radius if the motion is circular) are proportional according to the following relationship:
$$ T^2 \ \alpha \  R^3 $$
Call G the geostationary satellite and C the other.  Then
$$ \frac{T_G^2}{R_G^3} = \frac{T_C^2}{R_C^3} $$
Now you know that $$ T_G = 24 \ hours $$ precisely because G is geostationary.  From this, the other data in the question, and the relationship above, you can deduce the period of C, $ T_C $.  And then you can answer the question.

anonymous · Answer

yahhh thanks.....