A satelite revolves around a planet in an elliptical orbit. Its maximum and minimum distances from the planet are `1.5xx10^(7)` m and `0.5xx10^(7)` m respectively. If the speed of the satellite at the farthest point be `5xx10^(3)` m/s, calculate the speed at the nearest point.

Explanation

In case of elliptical orbit, the speed of satellite varies constantly as shown in figure. Thus according to the law of conservation of angular momentum, the satellite must move faster at a point of closest approach (Perigee) than at a farthest point (Appogee).
We know that, `vec(L) = vec(r ) xx m vec(v)`
Hence, at the two points,
`L = m v_(1)r_(1) = m v_(2)r_(2)`
or `(v_(1))/(v_(2)) = (r_(2))/(r_(1))`
Substituting the given values, we get
`(5 xx 10^(3))/(v_(2)) = (0.5 xx 10^(7))/(1.5 xx 10^(7))`
`rArr v_(2) = 1.5 xx 10^(4)` m/s