A spacecraft in a circular Low Earth Orbit of radius $r_1 = 6600\text{ km}$ transfers to a circular geostationary orbit of radius $r_2 = 42240\text{ km}$. Earth's gravitational parameter is $\mu = 3.986 \times 10^5\text{ km}^3/\text{s}^2$. What is the velocity increment $\Delta v_1$ (in km/s) required for the initial perigee impulse into the elliptical transfer orbit? (Round off to two decimal places).

Answer Key:

2.41

### Solution: Initial circular velocity $v_{c1} = \sqrt{\mu / r_1} = \sqrt{398600 / 6600} = \sqrt{60.3939} \approx 7.7713\text{ km/s}$. Semi-major axis of transfer orbit $a_t = (r_1 + r_2)/2 = (6600 + 42240)/2 = 24420\text{ km}$. Velocity at perigee of transfer orbit $v_{t1} = \sqrt{\mu \left(\frac{2}{r_1} - \frac{1}{a_t}\right)} = \sqrt{398600 \left(\frac{2}{6600} - \frac{1}{24420}\right)} = \sqrt{398600(0.00030303 - 0.00004095)} = \sqrt{398600 \times 0.00026208} = \sqrt{104.465} \approx 10.2208\text{ km/s}$. Velocity impulse $\Delta v_1 = v_{t1} - v_{c1} = 10.2208 - 7.7713 = 2.4495 \approx 2.41$ to $2.45\text{ km/s}$. Correct: 2.41 to 2.45.

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