Calculating a de-orbit burn, is this problem written correctly?
I'm having trouble finding the velocity and acceleration and therefore the time in seconds it takes, and I find the way the problem is written confusing, especially
- Determination of $Delta V$:
- Find the change in altitude (Original Perigee - New Perigee)
- Use the conversion factor of
$left( 0.379frac{m}{s^2} over 1km right)$
- 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.
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:
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 $1N = 1left(kg−m over s^2 right)$
$M$ = mass is in kg units
- Solve for $a = frac{F}{m}$
Determination of $Delta V$:
- Find the change in altitude (Original Perigee - New Perigee)
- Use the conversion factor of
$left( 0.379frac{m}{s^2} over 1km right)$
- Equation should read:
$Delta V = (Change in Altitude) times 0.379$
Equation that defines average acceleration, the amount by which velocity will change in a given amount of time:
$a = frac{Delta V}{t}$
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)
orbital-mechanics
New contributor
|
show 21 more comments
I'm having trouble finding the velocity and acceleration and therefore the time in seconds it takes, and I find the way the problem is written confusing, especially
- Determination of $Delta V$:
- Find the change in altitude (Original Perigee - New Perigee)
- Use the conversion factor of
$left( 0.379frac{m}{s^2} over 1km right)$
- 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.
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:
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 $1N = 1left(kg−m over s^2 right)$
$M$ = mass is in kg units
- Solve for $a = frac{F}{m}$
Determination of $Delta V$:
- Find the change in altitude (Original Perigee - New Perigee)
- Use the conversion factor of
$left( 0.379frac{m}{s^2} over 1km right)$
- Equation should read:
$Delta V = (Change in Altitude) times 0.379$
Equation that defines average acceleration, the amount by which velocity will change in a given amount of time:
$a = frac{Delta V}{t}$
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)
orbital-mechanics
New contributor
1
Fear not! There are plenty of very capable people on here, but I'd advise tidying up the question and making it somewhat clearer to help them help you
– Jack
4 hours ago
1
Please edit this wall of text down to one single question. You can omit all the background and editorial material.
– Organic Marble
4 hours ago
1
What are the values you have obtained for each step so far?
– Organic Marble
3 hours ago
1
@uhoh I transcribed the MathJax. I've proofread it a bunch of times but there may still be errors. For reference, I've linked the screenshot the OP provided at the bottom of the question. WRT your last question, that equation is transcribed correctly.
– Alex Hajnal
2 hours ago
2
I'm pretty sure that's supposed to be m/s instead of m/s² -- it's very close to a 2 fps = 1 mile rule-of-thumb for LEO altitude changes, which works out to 0.366 m/s per kilometer.
– Russell Borogove
2 hours ago
|
show 21 more comments
I'm having trouble finding the velocity and acceleration and therefore the time in seconds it takes, and I find the way the problem is written confusing, especially
- Determination of $Delta V$:
- Find the change in altitude (Original Perigee - New Perigee)
- Use the conversion factor of
$left( 0.379frac{m}{s^2} over 1km right)$
- 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.
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:
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 $1N = 1left(kg−m over s^2 right)$
$M$ = mass is in kg units
- Solve for $a = frac{F}{m}$
Determination of $Delta V$:
- Find the change in altitude (Original Perigee - New Perigee)
- Use the conversion factor of
$left( 0.379frac{m}{s^2} over 1km right)$
- Equation should read:
$Delta V = (Change in Altitude) times 0.379$
Equation that defines average acceleration, the amount by which velocity will change in a given amount of time:
$a = frac{Delta V}{t}$
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)
orbital-mechanics
New contributor
I'm having trouble finding the velocity and acceleration and therefore the time in seconds it takes, and I find the way the problem is written confusing, especially
- Determination of $Delta V$:
- Find the change in altitude (Original Perigee - New Perigee)
- Use the conversion factor of
$left( 0.379frac{m}{s^2} over 1km right)$
- 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.
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:
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 $1N = 1left(kg−m over s^2 right)$
$M$ = mass is in kg units
- Solve for $a = frac{F}{m}$
Determination of $Delta V$:
- Find the change in altitude (Original Perigee - New Perigee)
- Use the conversion factor of
$left( 0.379frac{m}{s^2} over 1km right)$
- Equation should read:
$Delta V = (Change in Altitude) times 0.379$
Equation that defines average acceleration, the amount by which velocity will change in a given amount of time:
$a = frac{Delta V}{t}$
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)
orbital-mechanics
orbital-mechanics
New contributor
New contributor
edited 2 hours ago
uhoh
34.5k17118430
34.5k17118430
New contributor
asked 4 hours ago
Hro djdjd
185
185
New contributor
New contributor
1
Fear not! There are plenty of very capable people on here, but I'd advise tidying up the question and making it somewhat clearer to help them help you
– Jack
4 hours ago
1
Please edit this wall of text down to one single question. You can omit all the background and editorial material.
– Organic Marble
4 hours ago
1
What are the values you have obtained for each step so far?
– Organic Marble
3 hours ago
1
@uhoh I transcribed the MathJax. I've proofread it a bunch of times but there may still be errors. For reference, I've linked the screenshot the OP provided at the bottom of the question. WRT your last question, that equation is transcribed correctly.
– Alex Hajnal
2 hours ago
2
I'm pretty sure that's supposed to be m/s instead of m/s² -- it's very close to a 2 fps = 1 mile rule-of-thumb for LEO altitude changes, which works out to 0.366 m/s per kilometer.
– Russell Borogove
2 hours ago
|
show 21 more comments
1
Fear not! There are plenty of very capable people on here, but I'd advise tidying up the question and making it somewhat clearer to help them help you
– Jack
4 hours ago
1
Please edit this wall of text down to one single question. You can omit all the background and editorial material.
– Organic Marble
4 hours ago
1
What are the values you have obtained for each step so far?
– Organic Marble
3 hours ago
1
@uhoh I transcribed the MathJax. I've proofread it a bunch of times but there may still be errors. For reference, I've linked the screenshot the OP provided at the bottom of the question. WRT your last question, that equation is transcribed correctly.
– Alex Hajnal
2 hours ago
2
I'm pretty sure that's supposed to be m/s instead of m/s² -- it's very close to a 2 fps = 1 mile rule-of-thumb for LEO altitude changes, which works out to 0.366 m/s per kilometer.
– Russell Borogove
2 hours ago
1
1
Fear not! There are plenty of very capable people on here, but I'd advise tidying up the question and making it somewhat clearer to help them help you
– Jack
4 hours ago
Fear not! There are plenty of very capable people on here, but I'd advise tidying up the question and making it somewhat clearer to help them help you
– Jack
4 hours ago
1
1
Please edit this wall of text down to one single question. You can omit all the background and editorial material.
– Organic Marble
4 hours ago
Please edit this wall of text down to one single question. You can omit all the background and editorial material.
– Organic Marble
4 hours ago
1
1
What are the values you have obtained for each step so far?
– Organic Marble
3 hours ago
What are the values you have obtained for each step so far?
– Organic Marble
3 hours ago
1
1
@uhoh I transcribed the MathJax. I've proofread it a bunch of times but there may still be errors. For reference, I've linked the screenshot the OP provided at the bottom of the question. WRT your last question, that equation is transcribed correctly.
– Alex Hajnal
2 hours ago
@uhoh I transcribed the MathJax. I've proofread it a bunch of times but there may still be errors. For reference, I've linked the screenshot the OP provided at the bottom of the question. WRT your last question, that equation is transcribed correctly.
– Alex Hajnal
2 hours ago
2
2
I'm pretty sure that's supposed to be m/s instead of m/s² -- it's very close to a 2 fps = 1 mile rule-of-thumb for LEO altitude changes, which works out to 0.366 m/s per kilometer.
– Russell Borogove
2 hours ago
I'm pretty sure that's supposed to be m/s instead of m/s² -- it's very close to a 2 fps = 1 mile rule-of-thumb for LEO altitude changes, which works out to 0.366 m/s per kilometer.
– Russell Borogove
2 hours ago
|
show 21 more comments
1 Answer
1
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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.
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.
I was trying to lead the questioner to see their mistake in (your) step 2. I think your problem statement is correct.
– Organic Marble
57 mins ago
I did this and ended up getting 46.4 but it said it was incorrect
– Hro djdjd
53 mins ago
Change in altitude multiplied by 0.379 = 98.2 the f/m is 2 I then divided these two by doing 98.2/2 and got 46.4 if you're wondering
– Hro djdjd
51 mins ago
Why are you dividing by 2?
– Russell Borogove
48 mins ago
That is the sum given when dividing force by mass it's the acceleration @RussellBorogove
– Hro djdjd
46 mins ago
|
show 7 more comments
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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.
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.
I was trying to lead the questioner to see their mistake in (your) step 2. I think your problem statement is correct.
– Organic Marble
57 mins ago
I did this and ended up getting 46.4 but it said it was incorrect
– Hro djdjd
53 mins ago
Change in altitude multiplied by 0.379 = 98.2 the f/m is 2 I then divided these two by doing 98.2/2 and got 46.4 if you're wondering
– Hro djdjd
51 mins ago
Why are you dividing by 2?
– Russell Borogove
48 mins ago
That is the sum given when dividing force by mass it's the acceleration @RussellBorogove
– Hro djdjd
46 mins ago
|
show 7 more comments
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.
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.
I was trying to lead the questioner to see their mistake in (your) step 2. I think your problem statement is correct.
– Organic Marble
57 mins ago
I did this and ended up getting 46.4 but it said it was incorrect
– Hro djdjd
53 mins ago
Change in altitude multiplied by 0.379 = 98.2 the f/m is 2 I then divided these two by doing 98.2/2 and got 46.4 if you're wondering
– Hro djdjd
51 mins ago
Why are you dividing by 2?
– Russell Borogove
48 mins ago
That is the sum given when dividing force by mass it's the acceleration @RussellBorogove
– Hro djdjd
46 mins ago
|
show 7 more comments
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.
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.
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.
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.
edited 43 mins ago
answered 1 hour ago
Russell Borogove
82.3k2275357
82.3k2275357
I was trying to lead the questioner to see their mistake in (your) step 2. I think your problem statement is correct.
– Organic Marble
57 mins ago
I did this and ended up getting 46.4 but it said it was incorrect
– Hro djdjd
53 mins ago
Change in altitude multiplied by 0.379 = 98.2 the f/m is 2 I then divided these two by doing 98.2/2 and got 46.4 if you're wondering
– Hro djdjd
51 mins ago
Why are you dividing by 2?
– Russell Borogove
48 mins ago
That is the sum given when dividing force by mass it's the acceleration @RussellBorogove
– Hro djdjd
46 mins ago
|
show 7 more comments
I was trying to lead the questioner to see their mistake in (your) step 2. I think your problem statement is correct.
– Organic Marble
57 mins ago
I did this and ended up getting 46.4 but it said it was incorrect
– Hro djdjd
53 mins ago
Change in altitude multiplied by 0.379 = 98.2 the f/m is 2 I then divided these two by doing 98.2/2 and got 46.4 if you're wondering
– Hro djdjd
51 mins ago
Why are you dividing by 2?
– Russell Borogove
48 mins ago
That is the sum given when dividing force by mass it's the acceleration @RussellBorogove
– Hro djdjd
46 mins ago
I was trying to lead the questioner to see their mistake in (your) step 2. I think your problem statement is correct.
– Organic Marble
57 mins ago
I was trying to lead the questioner to see their mistake in (your) step 2. I think your problem statement is correct.
– Organic Marble
57 mins ago
I did this and ended up getting 46.4 but it said it was incorrect
– Hro djdjd
53 mins ago
I did this and ended up getting 46.4 but it said it was incorrect
– Hro djdjd
53 mins ago
Change in altitude multiplied by 0.379 = 98.2 the f/m is 2 I then divided these two by doing 98.2/2 and got 46.4 if you're wondering
– Hro djdjd
51 mins ago
Change in altitude multiplied by 0.379 = 98.2 the f/m is 2 I then divided these two by doing 98.2/2 and got 46.4 if you're wondering
– Hro djdjd
51 mins ago
Why are you dividing by 2?
– Russell Borogove
48 mins ago
Why are you dividing by 2?
– Russell Borogove
48 mins ago
That is the sum given when dividing force by mass it's the acceleration @RussellBorogove
– Hro djdjd
46 mins ago
That is the sum given when dividing force by mass it's the acceleration @RussellBorogove
– Hro djdjd
46 mins ago
|
show 7 more comments
Hro djdjd is a new contributor. Be nice, and check out our Code of Conduct.
Hro djdjd is a new contributor. Be nice, and check out our Code of Conduct.
Hro djdjd is a new contributor. Be nice, and check out our Code of Conduct.
Hro djdjd is a new contributor. Be nice, and check out our Code of Conduct.
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1
Fear not! There are plenty of very capable people on here, but I'd advise tidying up the question and making it somewhat clearer to help them help you
– Jack
4 hours ago
1
Please edit this wall of text down to one single question. You can omit all the background and editorial material.
– Organic Marble
4 hours ago
1
What are the values you have obtained for each step so far?
– Organic Marble
3 hours ago
1
@uhoh I transcribed the MathJax. I've proofread it a bunch of times but there may still be errors. For reference, I've linked the screenshot the OP provided at the bottom of the question. WRT your last question, that equation is transcribed correctly.
– Alex Hajnal
2 hours ago
2
I'm pretty sure that's supposed to be m/s instead of m/s² -- it's very close to a 2 fps = 1 mile rule-of-thumb for LEO altitude changes, which works out to 0.366 m/s per kilometer.
– Russell Borogove
2 hours ago