an earth satellite has an elliptical orbit with an eccentricity value, e=0.40 and on orbital period of 10 hours and 54.0 minutes. Find the height above the surface at its closest approach point.
T = 654 min * 60 s /min = 39240 s
G = 6.673 * 10^-11 N*m ²/kg ²
M = 5.98 * 10^24 kg
T ² = 4 π ² a ³ / (GM)
Rearrange:
a ³ = T ² GM / (4 π ²)
Substitute the above values of T, G, amd M, and solve for a.
e = 0.40
Number of m above earth's surface at closest approach point is:
Rp = a (1 - e) - radius of earth
Substitute a, e, and earth's radius, and solve for Rp.