Calculating a de-orbit burn, is this problem written correctly?

Question

I'm having trouble finding the velocity and acceleration for deorbiting burn and therefore the time in seconds it takes, and I find the way the problem is written confusing, especially

2. Determination of $\Delta V$:

• Find the change in altitude (Original Perigee - New Perigee)
• Use the conversion factor of $\left( 0.379\frac{m}{s^2} \over 1km \right)$
• Equation should read: $\Delta V = (Change\ in\ Altitude) \times 0.379$

where the units do not even appear to agree, giving delta-v in m/s². Is it just me, or is there something amiss in the question?

Here is a screenshot of the original question, below has been kindly transcribed in edits.

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The question:

During a de-orbit burn, a pre-calculated ∆V (delta V, change in velocity) will be used to decrease the Orion MPCV’s altitude. The Orion MPCV’s Orbital Maneuvering System (OMS) engines provide a combined thrust force of 53,000 Newtons. The Orion MPCV has a mass of 25,848 kg when fully loaded.

What is the difference between the Orion MPCV’s mass and weight? An object’s mass does not change from place to place, but an object’s weight does change as it moves to a place with a different gravitational potential. For example, an object on the moon has the same mass it had while on the Earth but the object will weigh less on the moon due to the moon’s decreased gravitational potential. The Orion MPCV always has the same mass but will weigh less while in orbit than it does while on Earth’s surface.

CALCULATION: Calculate how long a de-orbit burn must last in seconds to achieve the Orion MPCV’s change in altitude from 343.5 kilometers to 96.5 kilometers at perigee. Use the equations and conversions provided to find the required burn time.

Equations to use:

1. Newton’s Second Law: $F=ma$

• Where:
• $a$ = acceleration is in meters per second per second $\left( m \over s^2 \right)$ units
• $F$ = force is in Newtons $1N = 1\left(kg−m \over s^2 \right)$
• $M$ = mass is in kg units
• Solve for $a = \frac{F}{m}$

2. Determination of $\Delta V$:

• Find the change in altitude (Original Perigee - New Perigee)
• Use the conversion factor of $\left( 0.379\frac{m}{s^2} \over 1km \right)$
• Equation should read: $\Delta V = (Change\ in\ Altitude) \times 0.379$

3. Equation that defines average acceleration, the amount by which velocity will change in a given amount of time: $a = \frac{\Delta V}{t}$

4. Rearranging the acceleration equation above to find the time required for a specific velocity change given a specific acceleration, where $t = \frac{\Delta V}{a}$

• $\Delta V$ = change in velocity in meters per second $m \over s$
• $a$ = acceleration is in meters per second per second, $m \over s^2$
• $t$ = required time in seconds (this is the value that you are solving for)

(The mix of $M$ and $m$ for both meters and mass is in the original text)

Answer 1

Assume that the 0.379 m/s² / km is a unit error, and the factor is supposed to be 0.379 m/s / km. I believe this is the fundamental mistake in the problem statement.

Step 1: The delta-v required is equal to the change in altitude in km, multiplied by the conversion factor. ∆v is measured in meters per second.

Step 2: Compute the acceleration of the spacecraft in m/s² by dividing the given thrust in N by the given mass in kg.

Step 3: Divide the ∆v by the acceleration to get time in seconds.

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The 0.379 m/s per km is a metricization of a rule of thumb in round Imperial-unit numbers: it's equal to 2 feet-per-second per mile, which isn't even particularly close to the correct value, which varies depending on the starting altitude but is generally closer to 0.3 for LEO maneuvers.