Question

An earth satellite moves in an elliptical orbit with a perigee altitude of 300 km and an apogee altitude of 3000 km. Assume that the radius of the earth is 6378 km. The eccentricity of the orbit is ________ (rounded off to three decimal places).

Hint:

Convert perigee and apogee altitudes to radii from earth's center, then use e = (ra - rp)/(ra + rp).

Answer 1

Correct Answer: 0.168

Step 1: Find the perigee and apogee radii from the earth's center.
Altitude is measured from the earth's surface, so add the earth's radius:
\[ r_p = R_e + h_p = 6378+300 = 6678\ \text{km} \]
\[ r_a = R_e + h_a = 6378+3000 = 9378\ \text{km} \]

Step 2: Recall the perigee/apogee relations for an ellipse.
For an orbit of semi-major axis \(a\) and eccentricity \(e\),
\[ r_p = a(1-e), \qquad r_a = a(1+e) \]

Step 3: Eliminate \(a\) to get \(e\) directly.
Dividing and rearranging,
\[ e = \frac{r_a-r_p}{r_a+r_p} \]

Step 4: Substitute the numbers.
\[ e = \frac{9378-6678}{9378+6678} = \frac{2700}{16056} = 0.1682 \]

Final Answer:
Rounded to three decimal places, \(e \approx 0.168\).
\[ \boxed{e \approx 0.168} \]