Compute the velocity required from Earth's surface to reach a circular orbit at 250 km altitude (round off to two decimals).
Earth data: \( GM_e = 398600.4 \, \text{km}^3/\text{s}^2 \), \( R_0 = 6378.14 \, \text{km} \)
Earth data: \( GM_e = 398600.4 \, \text{km}^3/\text{s}^2 \), \( R_0 = 6378.14 \, \text{km} \)
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Correct Answer: 7.75
Solution and Explanation
\[ r = R_0 + 250 = 6628.14 \, \text{km} \] Circular orbit velocity:
\[ v_\text{orbit} = \sqrt{\frac{GM_e}{r}} = \sqrt{\frac{398600.4}{6628.14}} \] Compute:
\[ v_\text{orbit} = \sqrt{60.09} = 7.75 \, \text{km/s} \] If converted to m/s: \[ 7.75 \text{ km/s} = 7750 \text{ m/s} \]
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