QuestionDecember 15, 2025

How long (in seconds)must a de-orbit burn last if the rocket is 25,848kg and a rocket force of 53,000 newtons to go from 334.8 km to 96.5 km if the mass does not change

How long (in seconds)must a de-orbit burn last if the rocket is 25,848kg and a rocket force of 53,000 newtons to go from 334.8 km to 96.5 km if the mass does not change
How long (in seconds)must a de-orbit burn last if the rocket is 25,848kg and a rocket force of
53,000 newtons to go from 334.8 km to 96.5 km if the mass does not change

Solution
4.1(407 votes)

Answer

56.6 seconds Explanation 1. Calculate change in velocity (Δv) Use v = \sqrt{GM/R} for circular orbits. R_1 = 6371 + 334.8 = 6705.8 km, R_2 = 6371 + 96.5 = 6467.5 km. GM = 3.986 \times 10^{14} \text{ m}^3/\text{s}^2. \Delta v = v_1 - v_2 = \sqrt{\frac{3.986 \times 10^{14}}{6.7058 \times 10^6}} - \sqrt{\frac{3.986 \times 10^{14}}{6.4675 \times 10^6}} v_1 \approx 7724.6 m/s, v_2 \approx 7840.7 m/s, so \Delta v = 7724.6 - 7840.7 = -116.1 m/s (negative means slowing down). 2. Find acceleration Use F = ma \implies a = F/m = 53000/25848 \approx 2.05 m/s². 3. Calculate burn time Use t = \Delta v / a = 116.1 / 2.05 \approx 56.6 s.

Explanation

1. Calculate change in velocity (Δv)<br /> Use $v = \sqrt{GM/R}$ for circular orbits. $R_1 = 6371 + 334.8 = 6705.8$ km, $R_2 = 6371 + 96.5 = 6467.5$ km. $GM = 3.986 \times 10^{14} \text{ m}^3/\text{s}^2$.<br /> $\Delta v = v_1 - v_2 = \sqrt{\frac{3.986 \times 10^{14}}{6.7058 \times 10^6}} - \sqrt{\frac{3.986 \times 10^{14}}{6.4675 \times 10^6}}$<br /> $v_1 \approx 7724.6$ m/s, $v_2 \approx 7840.7$ m/s, so $\Delta v = 7724.6 - 7840.7 = -116.1$ m/s (negative means slowing down).<br /><br />2. Find acceleration<br /> Use $F = ma \implies a = F/m = 53000/25848 \approx 2.05$ m/s².<br /><br />3. Calculate burn time<br /> Use $t = \Delta v / a = 116.1 / 2.05 \approx 56.6$ s.
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