The average radius of an ellipse [closed]
$begingroup$
I have recently been studying the 3rd Kepler´s law which uses the average radius of an orbit. I would love to know how to find the geometric average radius of an ellipse. Would you please show me or at least help me with the process of integration?
integration geometry
$endgroup$
closed as unclear what you're asking by José Carlos Santos, mrtaurho, abc..., Lee David Chung Lin, Thomas Shelby Feb 10 at 15:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have recently been studying the 3rd Kepler´s law which uses the average radius of an orbit. I would love to know how to find the geometric average radius of an ellipse. Would you please show me or at least help me with the process of integration?
integration geometry
$endgroup$
closed as unclear what you're asking by José Carlos Santos, mrtaurho, abc..., Lee David Chung Lin, Thomas Shelby Feb 10 at 15:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Please refer to another answer of mine here
$endgroup$
– Ng Chung Tak
Feb 10 at 10:37
3
$begingroup$
Beware: in the context of books presenting Kepler's laws, by "average radius" of an elliptic orbit they mean the semi-major axis, that is the average between maximum and minimum distance of a planet from the Sun. That is not what I would call "average radius" from a mathematical point of view.
$endgroup$
– Aretino
Feb 10 at 11:09
1
$begingroup$
@Aretino But I find it interesting that the average distance weighted by arc length is the same as the semi-major axis.
$endgroup$
– David K
Feb 10 at 15:47
add a comment |
$begingroup$
I have recently been studying the 3rd Kepler´s law which uses the average radius of an orbit. I would love to know how to find the geometric average radius of an ellipse. Would you please show me or at least help me with the process of integration?
integration geometry
$endgroup$
I have recently been studying the 3rd Kepler´s law which uses the average radius of an orbit. I would love to know how to find the geometric average radius of an ellipse. Would you please show me or at least help me with the process of integration?
integration geometry
integration geometry
edited Feb 10 at 15:23
Adam Páltik
asked Feb 10 at 9:34
Adam PáltikAdam Páltik
1149
1149
closed as unclear what you're asking by José Carlos Santos, mrtaurho, abc..., Lee David Chung Lin, Thomas Shelby Feb 10 at 15:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by José Carlos Santos, mrtaurho, abc..., Lee David Chung Lin, Thomas Shelby Feb 10 at 15:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Please refer to another answer of mine here
$endgroup$
– Ng Chung Tak
Feb 10 at 10:37
3
$begingroup$
Beware: in the context of books presenting Kepler's laws, by "average radius" of an elliptic orbit they mean the semi-major axis, that is the average between maximum and minimum distance of a planet from the Sun. That is not what I would call "average radius" from a mathematical point of view.
$endgroup$
– Aretino
Feb 10 at 11:09
1
$begingroup$
@Aretino But I find it interesting that the average distance weighted by arc length is the same as the semi-major axis.
$endgroup$
– David K
Feb 10 at 15:47
add a comment |
3
$begingroup$
Please refer to another answer of mine here
$endgroup$
– Ng Chung Tak
Feb 10 at 10:37
3
$begingroup$
Beware: in the context of books presenting Kepler's laws, by "average radius" of an elliptic orbit they mean the semi-major axis, that is the average between maximum and minimum distance of a planet from the Sun. That is not what I would call "average radius" from a mathematical point of view.
$endgroup$
– Aretino
Feb 10 at 11:09
1
$begingroup$
@Aretino But I find it interesting that the average distance weighted by arc length is the same as the semi-major axis.
$endgroup$
– David K
Feb 10 at 15:47
3
3
$begingroup$
Please refer to another answer of mine here
$endgroup$
– Ng Chung Tak
Feb 10 at 10:37
$begingroup$
Please refer to another answer of mine here
$endgroup$
– Ng Chung Tak
Feb 10 at 10:37
3
3
$begingroup$
Beware: in the context of books presenting Kepler's laws, by "average radius" of an elliptic orbit they mean the semi-major axis, that is the average between maximum and minimum distance of a planet from the Sun. That is not what I would call "average radius" from a mathematical point of view.
$endgroup$
– Aretino
Feb 10 at 11:09
$begingroup$
Beware: in the context of books presenting Kepler's laws, by "average radius" of an elliptic orbit they mean the semi-major axis, that is the average between maximum and minimum distance of a planet from the Sun. That is not what I would call "average radius" from a mathematical point of view.
$endgroup$
– Aretino
Feb 10 at 11:09
1
1
$begingroup$
@Aretino But I find it interesting that the average distance weighted by arc length is the same as the semi-major axis.
$endgroup$
– David K
Feb 10 at 15:47
$begingroup$
@Aretino But I find it interesting that the average distance weighted by arc length is the same as the semi-major axis.
$endgroup$
– David K
Feb 10 at 15:47
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hints:
The polar equation of the ellipse with one focus at the origin and the other at the positive real axis ($e = $ eccentricity) is:
$$r = frac{a(1 - e^2)}{1 - ecos(theta)}.$$
And the average radius can be calculated using the integral
$$frac{1}{2pi}int_0^{2pi}r(theta)dtheta.$$
Can you continue?
Update: already asked at Astronomy SE. The important fact:
It's the semi-major axis that defines the period, not the average distance.
Bottom line: the "simple average" $ne$ the integral average. Also interesting (and different): the time average.
Update 2: doing the cov $z = tan(theta/2)$,
$$intfrac{1}{1 - ecos(theta)}dtheta = intfrac{1}{1 - efrac{1 - z^2}{1 + z^2}}frac{2}{1 + z^2}dz = intfrac{2}{(1 + e)z^2 + (1 - e)}dz =$$
$$ = frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}zright) =
frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}tan(theta/2)right).$$
$endgroup$
$begingroup$
The result is $asqrt{1 - e^2}$, that is the semi-minor axis, but this is not what is meant by "average radius" in high-school books when presenting Kepler's third law.
$endgroup$
– Aretino
Feb 10 at 11:13
$begingroup$
@Aretino, see the update.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 11:45
$begingroup$
@Martín-BlasPérezPinilla How difficult is that integral to solve without the use of WolframAlpha?
$endgroup$
– Adam Páltik
Feb 10 at 15:19
$begingroup$
@AdamPáltik, the "universal change" $z = tan(x/2)$ (www-math.mit.edu/~djk/18_01/chapter24/section03.html) reduces it to a rational. I can do and post the calculations if you are interested.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 15:31
$begingroup$
@Martín-BlasPérezPinilla I would much appreciate that! :)
$endgroup$
– Adam Páltik
Feb 10 at 16:05
|
show 1 more comment
$begingroup$
@Martin-Blas has already suggested two possible notions of average
Take an average with respect to $theta$, the central angle
Take an average with respect to time, i.e.,$$
frac{1}{T}int_0^T | u(t) | dt,
$$
where $T$ is the period, and $u(t)$ is the position of the planet at time $t$, and the sun is located at the origin of the coordinate system.
There's a third notion, which is
- An average with respect to arclength, i.e., $$
int_0^L | u(t) | ds$$
where $L$ is the total arclength of the ellipse, and $ds = |u'(t)| dt$ is arclength integrand.
I'm pretty certain that one could even hoke up some others, but the key thing here is the one mentioned by others: this question doesn't have an answer until you know the measure with respect to which the "average" is being computed.
$endgroup$
add a comment |
$begingroup$
It depends on your definition of that average radius means in this context. E.g. if we define average radius of an ellipse as the radius $r$ of a circle which has the same area an an ellipse whose length of the semi-major and semi-minor axis are $a$ and $b$ then
$$
Area = pi r^2 = pi ab
$$
which gives $r = sqrt {ab}$.
Alternative definition is to literally take the average of $a$ and $b$ or $r = frac{a+b}{2}$.
$endgroup$
$begingroup$
Is it as simple as this?
$endgroup$
– Adam Páltik
Feb 10 at 10:04
4
$begingroup$
No. It is as simple as or as complicated as your definition of average. So in your question you need to clearly define what is the meaning of "average radius" in this context.
$endgroup$
– Nilotpal Kanti Sinha
Feb 10 at 10:06
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints:
The polar equation of the ellipse with one focus at the origin and the other at the positive real axis ($e = $ eccentricity) is:
$$r = frac{a(1 - e^2)}{1 - ecos(theta)}.$$
And the average radius can be calculated using the integral
$$frac{1}{2pi}int_0^{2pi}r(theta)dtheta.$$
Can you continue?
Update: already asked at Astronomy SE. The important fact:
It's the semi-major axis that defines the period, not the average distance.
Bottom line: the "simple average" $ne$ the integral average. Also interesting (and different): the time average.
Update 2: doing the cov $z = tan(theta/2)$,
$$intfrac{1}{1 - ecos(theta)}dtheta = intfrac{1}{1 - efrac{1 - z^2}{1 + z^2}}frac{2}{1 + z^2}dz = intfrac{2}{(1 + e)z^2 + (1 - e)}dz =$$
$$ = frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}zright) =
frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}tan(theta/2)right).$$
$endgroup$
$begingroup$
The result is $asqrt{1 - e^2}$, that is the semi-minor axis, but this is not what is meant by "average radius" in high-school books when presenting Kepler's third law.
$endgroup$
– Aretino
Feb 10 at 11:13
$begingroup$
@Aretino, see the update.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 11:45
$begingroup$
@Martín-BlasPérezPinilla How difficult is that integral to solve without the use of WolframAlpha?
$endgroup$
– Adam Páltik
Feb 10 at 15:19
$begingroup$
@AdamPáltik, the "universal change" $z = tan(x/2)$ (www-math.mit.edu/~djk/18_01/chapter24/section03.html) reduces it to a rational. I can do and post the calculations if you are interested.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 15:31
$begingroup$
@Martín-BlasPérezPinilla I would much appreciate that! :)
$endgroup$
– Adam Páltik
Feb 10 at 16:05
|
show 1 more comment
$begingroup$
Hints:
The polar equation of the ellipse with one focus at the origin and the other at the positive real axis ($e = $ eccentricity) is:
$$r = frac{a(1 - e^2)}{1 - ecos(theta)}.$$
And the average radius can be calculated using the integral
$$frac{1}{2pi}int_0^{2pi}r(theta)dtheta.$$
Can you continue?
Update: already asked at Astronomy SE. The important fact:
It's the semi-major axis that defines the period, not the average distance.
Bottom line: the "simple average" $ne$ the integral average. Also interesting (and different): the time average.
Update 2: doing the cov $z = tan(theta/2)$,
$$intfrac{1}{1 - ecos(theta)}dtheta = intfrac{1}{1 - efrac{1 - z^2}{1 + z^2}}frac{2}{1 + z^2}dz = intfrac{2}{(1 + e)z^2 + (1 - e)}dz =$$
$$ = frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}zright) =
frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}tan(theta/2)right).$$
$endgroup$
$begingroup$
The result is $asqrt{1 - e^2}$, that is the semi-minor axis, but this is not what is meant by "average radius" in high-school books when presenting Kepler's third law.
$endgroup$
– Aretino
Feb 10 at 11:13
$begingroup$
@Aretino, see the update.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 11:45
$begingroup$
@Martín-BlasPérezPinilla How difficult is that integral to solve without the use of WolframAlpha?
$endgroup$
– Adam Páltik
Feb 10 at 15:19
$begingroup$
@AdamPáltik, the "universal change" $z = tan(x/2)$ (www-math.mit.edu/~djk/18_01/chapter24/section03.html) reduces it to a rational. I can do and post the calculations if you are interested.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 15:31
$begingroup$
@Martín-BlasPérezPinilla I would much appreciate that! :)
$endgroup$
– Adam Páltik
Feb 10 at 16:05
|
show 1 more comment
$begingroup$
Hints:
The polar equation of the ellipse with one focus at the origin and the other at the positive real axis ($e = $ eccentricity) is:
$$r = frac{a(1 - e^2)}{1 - ecos(theta)}.$$
And the average radius can be calculated using the integral
$$frac{1}{2pi}int_0^{2pi}r(theta)dtheta.$$
Can you continue?
Update: already asked at Astronomy SE. The important fact:
It's the semi-major axis that defines the period, not the average distance.
Bottom line: the "simple average" $ne$ the integral average. Also interesting (and different): the time average.
Update 2: doing the cov $z = tan(theta/2)$,
$$intfrac{1}{1 - ecos(theta)}dtheta = intfrac{1}{1 - efrac{1 - z^2}{1 + z^2}}frac{2}{1 + z^2}dz = intfrac{2}{(1 + e)z^2 + (1 - e)}dz =$$
$$ = frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}zright) =
frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}tan(theta/2)right).$$
$endgroup$
Hints:
The polar equation of the ellipse with one focus at the origin and the other at the positive real axis ($e = $ eccentricity) is:
$$r = frac{a(1 - e^2)}{1 - ecos(theta)}.$$
And the average radius can be calculated using the integral
$$frac{1}{2pi}int_0^{2pi}r(theta)dtheta.$$
Can you continue?
Update: already asked at Astronomy SE. The important fact:
It's the semi-major axis that defines the period, not the average distance.
Bottom line: the "simple average" $ne$ the integral average. Also interesting (and different): the time average.
Update 2: doing the cov $z = tan(theta/2)$,
$$intfrac{1}{1 - ecos(theta)}dtheta = intfrac{1}{1 - efrac{1 - z^2}{1 + z^2}}frac{2}{1 + z^2}dz = intfrac{2}{(1 + e)z^2 + (1 - e)}dz =$$
$$ = frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}zright) =
frac{2}{sqrt{1 - e^2}}arctanleft(frac{sqrt{1 + e}}{sqrt{1 - e}}tan(theta/2)right).$$
edited Feb 11 at 9:54
answered Feb 10 at 10:50
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
35.4k42972
35.4k42972
$begingroup$
The result is $asqrt{1 - e^2}$, that is the semi-minor axis, but this is not what is meant by "average radius" in high-school books when presenting Kepler's third law.
$endgroup$
– Aretino
Feb 10 at 11:13
$begingroup$
@Aretino, see the update.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 11:45
$begingroup$
@Martín-BlasPérezPinilla How difficult is that integral to solve without the use of WolframAlpha?
$endgroup$
– Adam Páltik
Feb 10 at 15:19
$begingroup$
@AdamPáltik, the "universal change" $z = tan(x/2)$ (www-math.mit.edu/~djk/18_01/chapter24/section03.html) reduces it to a rational. I can do and post the calculations if you are interested.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 15:31
$begingroup$
@Martín-BlasPérezPinilla I would much appreciate that! :)
$endgroup$
– Adam Páltik
Feb 10 at 16:05
|
show 1 more comment
$begingroup$
The result is $asqrt{1 - e^2}$, that is the semi-minor axis, but this is not what is meant by "average radius" in high-school books when presenting Kepler's third law.
$endgroup$
– Aretino
Feb 10 at 11:13
$begingroup$
@Aretino, see the update.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 11:45
$begingroup$
@Martín-BlasPérezPinilla How difficult is that integral to solve without the use of WolframAlpha?
$endgroup$
– Adam Páltik
Feb 10 at 15:19
$begingroup$
@AdamPáltik, the "universal change" $z = tan(x/2)$ (www-math.mit.edu/~djk/18_01/chapter24/section03.html) reduces it to a rational. I can do and post the calculations if you are interested.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 15:31
$begingroup$
@Martín-BlasPérezPinilla I would much appreciate that! :)
$endgroup$
– Adam Páltik
Feb 10 at 16:05
$begingroup$
The result is $asqrt{1 - e^2}$, that is the semi-minor axis, but this is not what is meant by "average radius" in high-school books when presenting Kepler's third law.
$endgroup$
– Aretino
Feb 10 at 11:13
$begingroup$
The result is $asqrt{1 - e^2}$, that is the semi-minor axis, but this is not what is meant by "average radius" in high-school books when presenting Kepler's third law.
$endgroup$
– Aretino
Feb 10 at 11:13
$begingroup$
@Aretino, see the update.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 11:45
$begingroup$
@Aretino, see the update.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 11:45
$begingroup$
@Martín-BlasPérezPinilla How difficult is that integral to solve without the use of WolframAlpha?
$endgroup$
– Adam Páltik
Feb 10 at 15:19
$begingroup$
@Martín-BlasPérezPinilla How difficult is that integral to solve without the use of WolframAlpha?
$endgroup$
– Adam Páltik
Feb 10 at 15:19
$begingroup$
@AdamPáltik, the "universal change" $z = tan(x/2)$ (www-math.mit.edu/~djk/18_01/chapter24/section03.html) reduces it to a rational. I can do and post the calculations if you are interested.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 15:31
$begingroup$
@AdamPáltik, the "universal change" $z = tan(x/2)$ (www-math.mit.edu/~djk/18_01/chapter24/section03.html) reduces it to a rational. I can do and post the calculations if you are interested.
$endgroup$
– Martín-Blas Pérez Pinilla
Feb 10 at 15:31
$begingroup$
@Martín-BlasPérezPinilla I would much appreciate that! :)
$endgroup$
– Adam Páltik
Feb 10 at 16:05
$begingroup$
@Martín-BlasPérezPinilla I would much appreciate that! :)
$endgroup$
– Adam Páltik
Feb 10 at 16:05
|
show 1 more comment
$begingroup$
@Martin-Blas has already suggested two possible notions of average
Take an average with respect to $theta$, the central angle
Take an average with respect to time, i.e.,$$
frac{1}{T}int_0^T | u(t) | dt,
$$
where $T$ is the period, and $u(t)$ is the position of the planet at time $t$, and the sun is located at the origin of the coordinate system.
There's a third notion, which is
- An average with respect to arclength, i.e., $$
int_0^L | u(t) | ds$$
where $L$ is the total arclength of the ellipse, and $ds = |u'(t)| dt$ is arclength integrand.
I'm pretty certain that one could even hoke up some others, but the key thing here is the one mentioned by others: this question doesn't have an answer until you know the measure with respect to which the "average" is being computed.
$endgroup$
add a comment |
$begingroup$
@Martin-Blas has already suggested two possible notions of average
Take an average with respect to $theta$, the central angle
Take an average with respect to time, i.e.,$$
frac{1}{T}int_0^T | u(t) | dt,
$$
where $T$ is the period, and $u(t)$ is the position of the planet at time $t$, and the sun is located at the origin of the coordinate system.
There's a third notion, which is
- An average with respect to arclength, i.e., $$
int_0^L | u(t) | ds$$
where $L$ is the total arclength of the ellipse, and $ds = |u'(t)| dt$ is arclength integrand.
I'm pretty certain that one could even hoke up some others, but the key thing here is the one mentioned by others: this question doesn't have an answer until you know the measure with respect to which the "average" is being computed.
$endgroup$
add a comment |
$begingroup$
@Martin-Blas has already suggested two possible notions of average
Take an average with respect to $theta$, the central angle
Take an average with respect to time, i.e.,$$
frac{1}{T}int_0^T | u(t) | dt,
$$
where $T$ is the period, and $u(t)$ is the position of the planet at time $t$, and the sun is located at the origin of the coordinate system.
There's a third notion, which is
- An average with respect to arclength, i.e., $$
int_0^L | u(t) | ds$$
where $L$ is the total arclength of the ellipse, and $ds = |u'(t)| dt$ is arclength integrand.
I'm pretty certain that one could even hoke up some others, but the key thing here is the one mentioned by others: this question doesn't have an answer until you know the measure with respect to which the "average" is being computed.
$endgroup$
@Martin-Blas has already suggested two possible notions of average
Take an average with respect to $theta$, the central angle
Take an average with respect to time, i.e.,$$
frac{1}{T}int_0^T | u(t) | dt,
$$
where $T$ is the period, and $u(t)$ is the position of the planet at time $t$, and the sun is located at the origin of the coordinate system.
There's a third notion, which is
- An average with respect to arclength, i.e., $$
int_0^L | u(t) | ds$$
where $L$ is the total arclength of the ellipse, and $ds = |u'(t)| dt$ is arclength integrand.
I'm pretty certain that one could even hoke up some others, but the key thing here is the one mentioned by others: this question doesn't have an answer until you know the measure with respect to which the "average" is being computed.
answered Feb 10 at 12:26
John HughesJohn Hughes
65.4k24293
65.4k24293
add a comment |
add a comment |
$begingroup$
It depends on your definition of that average radius means in this context. E.g. if we define average radius of an ellipse as the radius $r$ of a circle which has the same area an an ellipse whose length of the semi-major and semi-minor axis are $a$ and $b$ then
$$
Area = pi r^2 = pi ab
$$
which gives $r = sqrt {ab}$.
Alternative definition is to literally take the average of $a$ and $b$ or $r = frac{a+b}{2}$.
$endgroup$
$begingroup$
Is it as simple as this?
$endgroup$
– Adam Páltik
Feb 10 at 10:04
4
$begingroup$
No. It is as simple as or as complicated as your definition of average. So in your question you need to clearly define what is the meaning of "average radius" in this context.
$endgroup$
– Nilotpal Kanti Sinha
Feb 10 at 10:06
add a comment |
$begingroup$
It depends on your definition of that average radius means in this context. E.g. if we define average radius of an ellipse as the radius $r$ of a circle which has the same area an an ellipse whose length of the semi-major and semi-minor axis are $a$ and $b$ then
$$
Area = pi r^2 = pi ab
$$
which gives $r = sqrt {ab}$.
Alternative definition is to literally take the average of $a$ and $b$ or $r = frac{a+b}{2}$.
$endgroup$
$begingroup$
Is it as simple as this?
$endgroup$
– Adam Páltik
Feb 10 at 10:04
4
$begingroup$
No. It is as simple as or as complicated as your definition of average. So in your question you need to clearly define what is the meaning of "average radius" in this context.
$endgroup$
– Nilotpal Kanti Sinha
Feb 10 at 10:06
add a comment |
$begingroup$
It depends on your definition of that average radius means in this context. E.g. if we define average radius of an ellipse as the radius $r$ of a circle which has the same area an an ellipse whose length of the semi-major and semi-minor axis are $a$ and $b$ then
$$
Area = pi r^2 = pi ab
$$
which gives $r = sqrt {ab}$.
Alternative definition is to literally take the average of $a$ and $b$ or $r = frac{a+b}{2}$.
$endgroup$
It depends on your definition of that average radius means in this context. E.g. if we define average radius of an ellipse as the radius $r$ of a circle which has the same area an an ellipse whose length of the semi-major and semi-minor axis are $a$ and $b$ then
$$
Area = pi r^2 = pi ab
$$
which gives $r = sqrt {ab}$.
Alternative definition is to literally take the average of $a$ and $b$ or $r = frac{a+b}{2}$.
edited Feb 11 at 4:31
answered Feb 10 at 9:59
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,67121641
4,67121641
$begingroup$
Is it as simple as this?
$endgroup$
– Adam Páltik
Feb 10 at 10:04
4
$begingroup$
No. It is as simple as or as complicated as your definition of average. So in your question you need to clearly define what is the meaning of "average radius" in this context.
$endgroup$
– Nilotpal Kanti Sinha
Feb 10 at 10:06
add a comment |
$begingroup$
Is it as simple as this?
$endgroup$
– Adam Páltik
Feb 10 at 10:04
4
$begingroup$
No. It is as simple as or as complicated as your definition of average. So in your question you need to clearly define what is the meaning of "average radius" in this context.
$endgroup$
– Nilotpal Kanti Sinha
Feb 10 at 10:06
$begingroup$
Is it as simple as this?
$endgroup$
– Adam Páltik
Feb 10 at 10:04
$begingroup$
Is it as simple as this?
$endgroup$
– Adam Páltik
Feb 10 at 10:04
4
4
$begingroup$
No. It is as simple as or as complicated as your definition of average. So in your question you need to clearly define what is the meaning of "average radius" in this context.
$endgroup$
– Nilotpal Kanti Sinha
Feb 10 at 10:06
$begingroup$
No. It is as simple as or as complicated as your definition of average. So in your question you need to clearly define what is the meaning of "average radius" in this context.
$endgroup$
– Nilotpal Kanti Sinha
Feb 10 at 10:06
add a comment |
3
$begingroup$
Please refer to another answer of mine here
$endgroup$
– Ng Chung Tak
Feb 10 at 10:37
3
$begingroup$
Beware: in the context of books presenting Kepler's laws, by "average radius" of an elliptic orbit they mean the semi-major axis, that is the average between maximum and minimum distance of a planet from the Sun. That is not what I would call "average radius" from a mathematical point of view.
$endgroup$
– Aretino
Feb 10 at 11:09
1
$begingroup$
@Aretino But I find it interesting that the average distance weighted by arc length is the same as the semi-major axis.
$endgroup$
– David K
Feb 10 at 15:47