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).

Question

A particle of mass $m$, and angular momentum $\ell$ is moving in a circular orbit of radius $r_0$ under the influence of an attractive force $\vec{F}(r) = -\frac{k}{r^2}\hat{r}$. Keeping its angular momentum unchanged, the particle is displaced radially by a small distance $\delta r \ll r_0$, due to which its radial distance varies periodically. The corresponding time period is:

Answer 1

Step 1: Find the semi-major axis first.
The semi-major axis is the average of perigee and apogee radii:
$a = \dfrac{r_p+r_a}{2} = \dfrac{6678+9378}{2} = \dfrac{16056}{2} = 8028$ km.

Step 2: Use the perigee relation to isolate $e$.
Since $r_p = a(1-e)$,
\[ e = 1 - \frac{r_p}{a} \]

Step 3: Substitute the numbers.
$e = 1 - \dfrac{6678}{8028} = 1-0.83184 = 0.16816$.

Step 4: Cross-check with the apogee relation.
Since $r_a = a(1+e)$, $e = r_a/a - 1 = 9378/8028 - 1 = 1.16816-1 = 0.16816$, which agrees with Step 3.

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

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).

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Correct Answer: 0.168

Solution and Explanation

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