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Why are the dimensions of escape velocity correct?
Clash Royale CLAN TAG#URR8PPP
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How does this formula work, from a dimensional analysis perspective?
$$ v_textescape = sqrtfrac2GMR$$
The way I'm thinking about it is that $G$ is in units $textN cdot textm^2/textkg^2$. You multiply by a kilogram amount (the mass) to turn $G$ into units $N cdot textm^2/textkg$. You then divide by the radius of the object to turn $G$ into units $N cdot textm/textkg$.
However, $v_textescape$ is in units $textm/texts$.
$sqrtN cdot textm/textkg neq textm/texts$.
Therefore, how does the equation even work if the units on either side aren't equal? Or am I doing this all wrong?
units dimensional-analysis escape-velocity
edited Aug 27 at 6:16
Nayuki
13026
asked Aug 26 at 7:33
virchau
343
 |Â
show 1 more comment
up vote
2
down vote
favorite
How does this formula work, from a dimensional analysis perspective?
$$ v_textescape = sqrtfrac2GMR$$
The way I'm thinking about it is that $G$ is in units $textN cdot textm^2/textkg^2$. You multiply by a kilogram amount (the mass) to turn $G$ into units $N cdot textm^2/textkg$. You then divide by the radius of the object to turn $G$ into units $N cdot textm/textkg$.
However, $v_textescape$ is in units $textm/texts$.
$sqrtN cdot textm/textkg neq textm/texts$.
Therefore, how does the equation even work if the units on either side aren't equal? Or am I doing this all wrong?
units dimensional-analysis escape-velocity
edited Aug 27 at 6:16
Nayuki
13026
asked Aug 26 at 7:33
virchau
343
Looks like an edit changed the meaning of the question, so it seems to answer itself... is a rollback in order?
– Chair
Aug 26 at 13:29
1
@Chair: I made the edit because I assumed that the problem was the meaning of $N$ as a unit, and that the loss of the square root was just a typo. Perhaps virchau could let us know?
– TonyK
Aug 26 at 13:31
1
@TonyK Yeah, it's best if we wait for some confirmation from virchau. I saw your description for the edit while reviewing it, but I'm inclined to believe that it's relatively hard to forget a square root mathjax command, considering the fact that it's a significant number of characters. That's the sticky thing about such questions...
– Chair
Aug 26 at 14:22
1
Yeah, agreed. I've put the question on hold until virchau can come back and clarify whether that missing square root was just a typo or if it was at the root of their confusion. (Note: a number of people thought this was a homework-like question, but at least in its current form, revision 5, it doesn't look like one to me.)
– David Z♦
Aug 27 at 6:54
@DavidZ: Sorry, I did indeed forget the square root. My main confusion was that I didn't realize that newtons aren't an SI unit; and Time4Tea's answer solved that confusion for me. This edit doesn't change the meaning of the question.
– virchau
Aug 28 at 13:26
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How does this formula work, from a dimensional analysis perspective?
$$ v_textescape = sqrtfrac2GMR$$
The way I'm thinking about it is that $G$ is in units $textN cdot textm^2/textkg^2$. You multiply by a kilogram amount (the mass) to turn $G$ into units $N cdot textm^2/textkg$. You then divide by the radius of the object to turn $G$ into units $N cdot textm/textkg$.
However, $v_textescape$ is in units $textm/texts$.
$sqrtN cdot textm/textkg neq textm/texts$.
Therefore, how does the equation even work if the units on either side aren't equal? Or am I doing this all wrong?
units dimensional-analysis escape-velocity
edited Aug 27 at 6:16
Nayuki
13026
asked Aug 26 at 7:33
virchau
343
How does this formula work, from a dimensional analysis perspective?
$$ v_textescape = sqrtfrac2GMR$$
The way I'm thinking about it is that $G$ is in units $textN cdot textm^2/textkg^2$. You multiply by a kilogram amount (the mass) to turn $G$ into units $N cdot textm^2/textkg$. You then divide by the radius of the object to turn $G$ into units $N cdot textm/textkg$.
However, $v_textescape$ is in units $textm/texts$.
$sqrtN cdot textm/textkg neq textm/texts$.
Therefore, how does the equation even work if the units on either side aren't equal? Or am I doing this all wrong?
units dimensional-analysis escape-velocity
edited Aug 27 at 6:16
Nayuki
13026
asked Aug 26 at 7:33
virchau
343
edited Aug 27 at 6:16
Nayuki
13026
edited Aug 27 at 6:16
Nayuki
13026
edited Aug 27 at 6:16
Nayuki
13026
13026
asked Aug 26 at 7:33
virchau
343
asked Aug 26 at 7:33
virchau
343
asked Aug 26 at 7:33
virchau
343
343
Looks like an edit changed the meaning of the question, so it seems to answer itself... is a rollback in order?
– Chair
Aug 26 at 13:29
1
@Chair: I made the edit because I assumed that the problem was the meaning of $N$ as a unit, and that the loss of the square root was just a typo. Perhaps virchau could let us know?
– TonyK
Aug 26 at 13:31
1
@TonyK Yeah, it's best if we wait for some confirmation from virchau. I saw your description for the edit while reviewing it, but I'm inclined to believe that it's relatively hard to forget a square root mathjax command, considering the fact that it's a significant number of characters. That's the sticky thing about such questions...
– Chair
Aug 26 at 14:22
1
Yeah, agreed. I've put the question on hold until virchau can come back and clarify whether that missing square root was just a typo or if it was at the root of their confusion. (Note: a number of people thought this was a homework-like question, but at least in its current form, revision 5, it doesn't look like one to me.)
– David Z♦
Aug 27 at 6:54
@DavidZ: Sorry, I did indeed forget the square root. My main confusion was that I didn't realize that newtons aren't an SI unit; and Time4Tea's answer solved that confusion for me. This edit doesn't change the meaning of the question.
– virchau
Aug 28 at 13:26
 |Â
show 1 more comment
Looks like an edit changed the meaning of the question, so it seems to answer itself... is a rollback in order?
– Chair
Aug 26 at 13:29
1
@Chair: I made the edit because I assumed that the problem was the meaning of $N$ as a unit, and that the loss of the square root was just a typo. Perhaps virchau could let us know?
– TonyK
Aug 26 at 13:31
1
@TonyK Yeah, it's best if we wait for some confirmation from virchau. I saw your description for the edit while reviewing it, but I'm inclined to believe that it's relatively hard to forget a square root mathjax command, considering the fact that it's a significant number of characters. That's the sticky thing about such questions...
– Chair
Aug 26 at 14:22
1
Yeah, agreed. I've put the question on hold until virchau can come back and clarify whether that missing square root was just a typo or if it was at the root of their confusion. (Note: a number of people thought this was a homework-like question, but at least in its current form, revision 5, it doesn't look like one to me.)
– David Z♦
Aug 27 at 6:54
@DavidZ: Sorry, I did indeed forget the square root. My main confusion was that I didn't realize that newtons aren't an SI unit; and Time4Tea's answer solved that confusion for me. This edit doesn't change the meaning of the question.
– virchau
Aug 28 at 13:26
Looks like an edit changed the meaning of the question, so it seems to answer itself... is a rollback in order?
– Chair
Aug 26 at 13:29
Looks like an edit changed the meaning of the question, so it seems to answer itself... is a rollback in order?
– Chair
Aug 26 at 13:29
1
1
@Chair: I made the edit because I assumed that the problem was the meaning of $N$ as a unit, and that the loss of the square root was just a typo. Perhaps virchau could let us know?
– TonyK
Aug 26 at 13:31
@Chair: I made the edit because I assumed that the problem was the meaning of $N$ as a unit, and that the loss of the square root was just a typo. Perhaps virchau could let us know?
– TonyK
Aug 26 at 13:31
1
1
@TonyK Yeah, it's best if we wait for some confirmation from virchau. I saw your description for the edit while reviewing it, but I'm inclined to believe that it's relatively hard to forget a square root mathjax command, considering the fact that it's a significant number of characters. That's the sticky thing about such questions...
– Chair
Aug 26 at 14:22
@TonyK Yeah, it's best if we wait for some confirmation from virchau. I saw your description for the edit while reviewing it, but I'm inclined to believe that it's relatively hard to forget a square root mathjax command, considering the fact that it's a significant number of characters. That's the sticky thing about such questions...
– Chair
Aug 26 at 14:22
1
1
Yeah, agreed. I've put the question on hold until virchau can come back and clarify whether that missing square root was just a typo or if it was at the root of their confusion. (Note: a number of people thought this was a homework-like question, but at least in its current form, revision 5, it doesn't look like one to me.)
– David Z♦
Aug 27 at 6:54
Yeah, agreed. I've put the question on hold until virchau can come back and clarify whether that missing square root was just a typo or if it was at the root of their confusion. (Note: a number of people thought this was a homework-like question, but at least in its current form, revision 5, it doesn't look like one to me.)
– David Z♦
Aug 27 at 6:54
@DavidZ: Sorry, I did indeed forget the square root. My main confusion was that I didn't realize that newtons aren't an SI unit; and Time4Tea's answer solved that confusion for me. This edit doesn't change the meaning of the question.
– virchau
Aug 28 at 13:26
@DavidZ: Sorry, I did indeed forget the square root. My main confusion was that I didn't realize that newtons aren't an SI unit; and Time4Tea's answer solved that confusion for me. This edit doesn't change the meaning of the question.
– virchau
Aug 28 at 13:26
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
24
down vote
accepted
Newton is not a fundamental SI unit:
$$mathrm N=fracmathrmkgcdotmathrm mmathrm s^2.$$
So, in fact:
$$fracmathrm Ncdotmathrm mmathrmkg=fracmathrm m^2mathrm s^2,$$
the square root of which has the units of velocity.
edited Aug 26 at 13:40
Ruslan
6,91342662
answered Aug 26 at 7:41
Time4Tea
2,131929
add a comment |Â
up vote
11
down vote
You forget that $mathrmN = mathrmkg mathrmm/mathrms^2.$
answered Aug 26 at 7:44
md2perpe
37425
How does that answer the question?
– Peter Mortensen
Aug 26 at 11:37
14
@PeterMortensen: it answers the question succinctly and accurately. It tells the OP that $sqrtN cdot m/kg$ is indeed equal to $m/s$.
– TonyK
Aug 26 at 13:01
1
It'd be better if you could expand this hint into a full-fledged answer.
– Chair
Aug 26 at 13:27
5
Since Time4Tea has said the same thing and that answer has been accepted, I think that there's currently no need to expand my answer.
– md2perpe
Aug 26 at 14:31
5
A few thoughts on context and audience. If a student came to my office hours with this question I would probably give them exactly this answer. Because it is better if they follow the logic through themselves. But I would also sit there paying close attention to their face while they worked on it. Because in the event that they are too confused to see the way forward my job isn't done at that point. I still have to prod or lead them forward until they get it. So whether this answer is complete or not depends on the reader. Just one of the oddities of Stack Exchange.
– dmckee♦
Aug 27 at 2:27
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
24
down vote
accepted
Newton is not a fundamental SI unit:
$$mathrm N=fracmathrmkgcdotmathrm mmathrm s^2.$$
So, in fact:
$$fracmathrm Ncdotmathrm mmathrmkg=fracmathrm m^2mathrm s^2,$$
the square root of which has the units of velocity.
edited Aug 26 at 13:40
Ruslan
6,91342662
answered Aug 26 at 7:41
Time4Tea
2,131929
add a comment |Â
up vote
24
down vote
accepted
Newton is not a fundamental SI unit:
$$mathrm N=fracmathrmkgcdotmathrm mmathrm s^2.$$
So, in fact:
$$fracmathrm Ncdotmathrm mmathrmkg=fracmathrm m^2mathrm s^2,$$
the square root of which has the units of velocity.
edited Aug 26 at 13:40
Ruslan
6,91342662
answered Aug 26 at 7:41
Time4Tea
2,131929
add a comment |Â
up vote
24
down vote
accepted
up vote
24
down vote
accepted
Newton is not a fundamental SI unit:
$$mathrm N=fracmathrmkgcdotmathrm mmathrm s^2.$$
So, in fact:
$$fracmathrm Ncdotmathrm mmathrmkg=fracmathrm m^2mathrm s^2,$$
the square root of which has the units of velocity.
edited Aug 26 at 13:40
Ruslan
6,91342662
answered Aug 26 at 7:41
Time4Tea
2,131929
Newton is not a fundamental SI unit:
$$mathrm N=fracmathrmkgcdotmathrm mmathrm s^2.$$
So, in fact:
$$fracmathrm Ncdotmathrm mmathrmkg=fracmathrm m^2mathrm s^2,$$
the square root of which has the units of velocity.
edited Aug 26 at 13:40
Ruslan
6,91342662
answered Aug 26 at 7:41
Time4Tea
2,131929
edited Aug 26 at 13:40
Ruslan
6,91342662
edited Aug 26 at 13:40
Ruslan
6,91342662
edited Aug 26 at 13:40
Ruslan
6,91342662
6,91342662
answered Aug 26 at 7:41
Time4Tea
2,131929
answered Aug 26 at 7:41
Time4Tea
2,131929
answered Aug 26 at 7:41
Time4Tea
2,131929
2,131929
add a comment |Â
add a comment |Â
up vote
11
down vote
You forget that $mathrmN = mathrmkg mathrmm/mathrms^2.$
answered Aug 26 at 7:44
md2perpe
37425
How does that answer the question?
– Peter Mortensen
Aug 26 at 11:37
14
@PeterMortensen: it answers the question succinctly and accurately. It tells the OP that $sqrtN cdot m/kg$ is indeed equal to $m/s$.
– TonyK
Aug 26 at 13:01
1
It'd be better if you could expand this hint into a full-fledged answer.
– Chair
Aug 26 at 13:27
5
Since Time4Tea has said the same thing and that answer has been accepted, I think that there's currently no need to expand my answer.
– md2perpe
Aug 26 at 14:31
5
A few thoughts on context and audience. If a student came to my office hours with this question I would probably give them exactly this answer. Because it is better if they follow the logic through themselves. But I would also sit there paying close attention to their face while they worked on it. Because in the event that they are too confused to see the way forward my job isn't done at that point. I still have to prod or lead them forward until they get it. So whether this answer is complete or not depends on the reader. Just one of the oddities of Stack Exchange.
– dmckee♦
Aug 27 at 2:27
 |Â
show 1 more comment
up vote
11
down vote
You forget that $mathrmN = mathrmkg mathrmm/mathrms^2.$
answered Aug 26 at 7:44
md2perpe
37425
How does that answer the question?
– Peter Mortensen
Aug 26 at 11:37
14
@PeterMortensen: it answers the question succinctly and accurately. It tells the OP that $sqrtN cdot m/kg$ is indeed equal to $m/s$.
– TonyK
Aug 26 at 13:01
1
It'd be better if you could expand this hint into a full-fledged answer.
– Chair
Aug 26 at 13:27
5
Since Time4Tea has said the same thing and that answer has been accepted, I think that there's currently no need to expand my answer.
– md2perpe
Aug 26 at 14:31
5
A few thoughts on context and audience. If a student came to my office hours with this question I would probably give them exactly this answer. Because it is better if they follow the logic through themselves. But I would also sit there paying close attention to their face while they worked on it. Because in the event that they are too confused to see the way forward my job isn't done at that point. I still have to prod or lead them forward until they get it. So whether this answer is complete or not depends on the reader. Just one of the oddities of Stack Exchange.
– dmckee♦
Aug 27 at 2:27
 |Â
show 1 more comment
up vote
11
down vote
up vote
11
down vote
You forget that $mathrmN = mathrmkg mathrmm/mathrms^2.$
answered Aug 26 at 7:44
md2perpe
37425
You forget that $mathrmN = mathrmkg mathrmm/mathrms^2.$
answered Aug 26 at 7:44
md2perpe
37425
answered Aug 26 at 7:44
md2perpe
37425
answered Aug 26 at 7:44
md2perpe
37425
answered Aug 26 at 7:44
md2perpe
37425
37425
How does that answer the question?
– Peter Mortensen
Aug 26 at 11:37
14
@PeterMortensen: it answers the question succinctly and accurately. It tells the OP that $sqrtN cdot m/kg$ is indeed equal to $m/s$.
– TonyK
Aug 26 at 13:01
1
It'd be better if you could expand this hint into a full-fledged answer.
– Chair
Aug 26 at 13:27
5
Since Time4Tea has said the same thing and that answer has been accepted, I think that there's currently no need to expand my answer.
– md2perpe
Aug 26 at 14:31
5
A few thoughts on context and audience. If a student came to my office hours with this question I would probably give them exactly this answer. Because it is better if they follow the logic through themselves. But I would also sit there paying close attention to their face while they worked on it. Because in the event that they are too confused to see the way forward my job isn't done at that point. I still have to prod or lead them forward until they get it. So whether this answer is complete or not depends on the reader. Just one of the oddities of Stack Exchange.
– dmckee♦
Aug 27 at 2:27
 |Â
show 1 more comment
How does that answer the question?
– Peter Mortensen
Aug 26 at 11:37
14
@PeterMortensen: it answers the question succinctly and accurately. It tells the OP that $sqrtN cdot m/kg$ is indeed equal to $m/s$.
– TonyK
Aug 26 at 13:01
1
It'd be better if you could expand this hint into a full-fledged answer.
– Chair
Aug 26 at 13:27
5
Since Time4Tea has said the same thing and that answer has been accepted, I think that there's currently no need to expand my answer.
– md2perpe
Aug 26 at 14:31
5
A few thoughts on context and audience. If a student came to my office hours with this question I would probably give them exactly this answer. Because it is better if they follow the logic through themselves. But I would also sit there paying close attention to their face while they worked on it. Because in the event that they are too confused to see the way forward my job isn't done at that point. I still have to prod or lead them forward until they get it. So whether this answer is complete or not depends on the reader. Just one of the oddities of Stack Exchange.
– dmckee♦
Aug 27 at 2:27
How does that answer the question?
– Peter Mortensen
Aug 26 at 11:37
How does that answer the question?
– Peter Mortensen
Aug 26 at 11:37
14
14
@PeterMortensen: it answers the question succinctly and accurately. It tells the OP that $sqrtN cdot m/kg$ is indeed equal to $m/s$.
– TonyK
Aug 26 at 13:01
@PeterMortensen: it answers the question succinctly and accurately. It tells the OP that $sqrtN cdot m/kg$ is indeed equal to $m/s$.
– TonyK
Aug 26 at 13:01
1
1
It'd be better if you could expand this hint into a full-fledged answer.
– Chair
Aug 26 at 13:27
It'd be better if you could expand this hint into a full-fledged answer.
– Chair
Aug 26 at 13:27
5
5
Since Time4Tea has said the same thing and that answer has been accepted, I think that there's currently no need to expand my answer.
– md2perpe
Aug 26 at 14:31
Since Time4Tea has said the same thing and that answer has been accepted, I think that there's currently no need to expand my answer.
– md2perpe
Aug 26 at 14:31
5
5
A few thoughts on context and audience. If a student came to my office hours with this question I would probably give them exactly this answer. Because it is better if they follow the logic through themselves. But I would also sit there paying close attention to their face while they worked on it. Because in the event that they are too confused to see the way forward my job isn't done at that point. I still have to prod or lead them forward until they get it. So whether this answer is complete or not depends on the reader. Just one of the oddities of Stack Exchange.
– dmckee♦
Aug 27 at 2:27
A few thoughts on context and audience. If a student came to my office hours with this question I would probably give them exactly this answer. Because it is better if they follow the logic through themselves. But I would also sit there paying close attention to their face while they worked on it. Because in the event that they are too confused to see the way forward my job isn't done at that point. I still have to prod or lead them forward until they get it. So whether this answer is complete or not depends on the reader. Just one of the oddities of Stack Exchange.
– dmckee♦
Aug 27 at 2:27
 |Â
show 1 more comment
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Looks like an edit changed the meaning of the question, so it seems to answer itself... is a rollback in order?
– Chair
Aug 26 at 13:29
1
@Chair: I made the edit because I assumed that the problem was the meaning of $N$ as a unit, and that the loss of the square root was just a typo. Perhaps virchau could let us know?
– TonyK
Aug 26 at 13:31
1
@TonyK Yeah, it's best if we wait for some confirmation from virchau. I saw your description for the edit while reviewing it, but I'm inclined to believe that it's relatively hard to forget a square root mathjax command, considering the fact that it's a significant number of characters. That's the sticky thing about such questions...
– Chair
Aug 26 at 14:22
1
Yeah, agreed. I've put the question on hold until virchau can come back and clarify whether that missing square root was just a typo or if it was at the root of their confusion. (Note: a number of people thought this was a homework-like question, but at least in its current form, revision 5, it doesn't look like one to me.)
– David Z♦
Aug 27 at 6:54
@DavidZ: Sorry, I did indeed forget the square root. My main confusion was that I didn't realize that newtons aren't an SI unit; and Time4Tea's answer solved that confusion for me. This edit doesn't change the meaning of the question.
– virchau
Aug 28 at 13:26