Kepler's 3rd law: ratios don't fit data
$begingroup$
I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.
Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.
I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.
By solving the equation for $a$, I get $a = (P^2)^{1/3}$.
When I plug in the numbers, they don't correspond.
So my questions are:
- Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.
- Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.
newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites
edited 10 hours ago
Qmechanic♦
108k122001253
asked 11 hours ago
triple7triple7
233
New contributor
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 1 more comment
$begingroup$
I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.
Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.
I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.
By solving the equation for $a$, I get $a = (P^2)^{1/3}$.
When I plug in the numbers, they don't correspond.
So my questions are:
- Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.
- Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.
newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites
edited 10 hours ago
Qmechanic♦
108k122001253
asked 11 hours ago
triple7triple7
233
New contributor
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
11 hours ago
$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
8 hours ago
1
$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
5 hours ago
$begingroup$
No satellite orbits earth at an altitude of 705 of any commonly used unit--that's either so far out it's lost or it's deep under the surface. Kepler's third law is talking about the distance between the centers, not the distance between the surfaces.
$endgroup$
– Loren Pechtel
58 mins ago
$begingroup$
Aqua has a 95 minute orbit revolution and is at 705 km altitude above earth’s surface. I have taken into account earth’s radius...
$endgroup$
– triple7
42 mins ago
|
show 1 more comment
$begingroup$
I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.
Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.
I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.
By solving the equation for $a$, I get $a = (P^2)^{1/3}$.
When I plug in the numbers, they don't correspond.
So my questions are:
- Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.
- Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.
newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites
edited 10 hours ago
Qmechanic♦
108k122001253
asked 11 hours ago
triple7triple7
233
New contributor
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.
Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.
I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.
By solving the equation for $a$, I get $a = (P^2)^{1/3}$.
When I plug in the numbers, they don't correspond.
So my questions are:
- Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.
- Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.
newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites
newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites
edited 10 hours ago
Qmechanic♦
108k122001253
asked 11 hours ago
triple7triple7
233
New contributor
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
Qmechanic♦
108k122001253
asked 11 hours ago
triple7triple7
233
New contributor
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
Qmechanic♦
108k122001253
edited 10 hours ago
Qmechanic♦
108k122001253
edited 10 hours ago
Qmechanic♦
108k122001253
108k122001253
asked 11 hours ago
triple7triple7
233
New contributor
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
triple7triple7
233
asked 11 hours ago
triple7triple7
233
233
New contributor
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
11 hours ago
$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
8 hours ago
1
$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
5 hours ago
$begingroup$
No satellite orbits earth at an altitude of 705 of any commonly used unit--that's either so far out it's lost or it's deep under the surface. Kepler's third law is talking about the distance between the centers, not the distance between the surfaces.
$endgroup$
– Loren Pechtel
58 mins ago
$begingroup$
Aqua has a 95 minute orbit revolution and is at 705 km altitude above earth’s surface. I have taken into account earth’s radius...
$endgroup$
– triple7
42 mins ago
|
show 1 more comment
$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
11 hours ago
$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
8 hours ago
1
$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
5 hours ago
$begingroup$
No satellite orbits earth at an altitude of 705 of any commonly used unit--that's either so far out it's lost or it's deep under the surface. Kepler's third law is talking about the distance between the centers, not the distance between the surfaces.
$endgroup$
– Loren Pechtel
58 mins ago
$begingroup$
Aqua has a 95 minute orbit revolution and is at 705 km altitude above earth’s surface. I have taken into account earth’s radius...
$endgroup$
– triple7
42 mins ago
$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
11 hours ago
$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
11 hours ago
$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
8 hours ago
$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
8 hours ago
1
1
$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
5 hours ago
$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
5 hours ago
$begingroup$
No satellite orbits earth at an altitude of 705 of any commonly used unit--that's either so far out it's lost or it's deep under the surface. Kepler's third law is talking about the distance between the centers, not the distance between the surfaces.
$endgroup$
– Loren Pechtel
58 mins ago
$begingroup$
No satellite orbits earth at an altitude of 705 of any commonly used unit--that's either so far out it's lost or it's deep under the surface. Kepler's third law is talking about the distance between the centers, not the distance between the surfaces.
$endgroup$
– Loren Pechtel
58 mins ago
$begingroup$
Aqua has a 95 minute orbit revolution and is at 705 km altitude above earth’s surface. I have taken into account earth’s radius...
$endgroup$
– triple7
42 mins ago
$begingroup$
Aqua has a 95 minute orbit revolution and is at 705 km altitude above earth’s surface. I have taken into account earth’s radius...
$endgroup$
– triple7
42 mins ago
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.
Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)
answered 11 hours ago
Jerry SchirmerJerry Schirmer
31.7k257107
$endgroup$
6
$begingroup$
Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
$endgroup$
– triple7
11 hours ago
3
$begingroup$
@triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
$endgroup$
– Jerry Schirmer
10 hours ago
1
$begingroup$
Yep, finally found it. Thanks
$endgroup$
– triple7
10 hours ago
add a comment |
$begingroup$
The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.
Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.
Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!
answered 11 hours ago
swickrotationswickrotation
715
$endgroup$
1
$begingroup$
Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
$endgroup$
– triple7
1 hour ago
add a comment |
$begingroup$
Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.
answered 11 hours ago
my2ctsmy2cts
5,9642719
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.
Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)
answered 11 hours ago
Jerry SchirmerJerry Schirmer
31.7k257107
$endgroup$
6
$begingroup$
Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
$endgroup$
– triple7
11 hours ago
3
$begingroup$
@triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
$endgroup$
– Jerry Schirmer
10 hours ago
1
$begingroup$
Yep, finally found it. Thanks
$endgroup$
– triple7
10 hours ago
add a comment |
$begingroup$
that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.
Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)
answered 11 hours ago
Jerry SchirmerJerry Schirmer
31.7k257107
$endgroup$
6
$begingroup$
Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
$endgroup$
– triple7
11 hours ago
3
$begingroup$
@triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
$endgroup$
– Jerry Schirmer
10 hours ago
1
$begingroup$
Yep, finally found it. Thanks
$endgroup$
– triple7
10 hours ago
add a comment |
$begingroup$
that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.
Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)
answered 11 hours ago
Jerry SchirmerJerry Schirmer
31.7k257107
$endgroup$
that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.
Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)
answered 11 hours ago
Jerry SchirmerJerry Schirmer
31.7k257107
edited 10 hours ago
answered 11 hours ago
Jerry SchirmerJerry Schirmer
31.7k257107
answered 11 hours ago
Jerry SchirmerJerry Schirmer
31.7k257107
answered 11 hours ago
Jerry SchirmerJerry Schirmer
31.7k257107
31.7k257107
6
$begingroup$
Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
$endgroup$
– triple7
11 hours ago
3
$begingroup$
@triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
$endgroup$
– Jerry Schirmer
10 hours ago
1
$begingroup$
Yep, finally found it. Thanks
$endgroup$
– triple7
10 hours ago
add a comment |
6
$begingroup$
Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
$endgroup$
– triple7
11 hours ago
3
$begingroup$
@triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
$endgroup$
– Jerry Schirmer
10 hours ago
1
$begingroup$
Yep, finally found it. Thanks
$endgroup$
– triple7
10 hours ago
6
6
$begingroup$
Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
$endgroup$
– triple7
11 hours ago
$begingroup$
Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
$endgroup$
– triple7
11 hours ago
3
3
$begingroup$
@triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
$endgroup$
– Jerry Schirmer
10 hours ago
$begingroup$
@triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
$endgroup$
– Jerry Schirmer
10 hours ago
1
1
$begingroup$
Yep, finally found it. Thanks
$endgroup$
– triple7
10 hours ago
$begingroup$
Yep, finally found it. Thanks
$endgroup$
– triple7
10 hours ago
add a comment |
$begingroup$
The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.
Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.
Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!
answered 11 hours ago
swickrotationswickrotation
715
$endgroup$
1
$begingroup$
Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
$endgroup$
– triple7
1 hour ago
add a comment |
$begingroup$
The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.
Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.
Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!
answered 11 hours ago
swickrotationswickrotation
715
$endgroup$
1
$begingroup$
Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
$endgroup$
– triple7
1 hour ago
add a comment |
$begingroup$
The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.
Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.
Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!
answered 11 hours ago
swickrotationswickrotation
715
$endgroup$
The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.
Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.
Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!
answered 11 hours ago
swickrotationswickrotation
715
edited 10 hours ago
answered 11 hours ago
swickrotationswickrotation
715
answered 11 hours ago
swickrotationswickrotation
715
answered 11 hours ago
swickrotationswickrotation
715
715
1
$begingroup$
Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
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– triple7
1 hour ago
add a comment |
1
$begingroup$
Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
$endgroup$
– triple7
1 hour ago
1
1
$begingroup$
Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
$endgroup$
– triple7
1 hour ago
$begingroup$
Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
$endgroup$
– triple7
1 hour ago
add a comment |
$begingroup$
Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.
answered 11 hours ago
my2ctsmy2cts
5,9642719
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add a comment |
$begingroup$
Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.
answered 11 hours ago
my2ctsmy2cts
5,9642719
$endgroup$
add a comment |
$begingroup$
Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.
answered 11 hours ago
my2ctsmy2cts
5,9642719
$endgroup$
Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.
answered 11 hours ago
my2ctsmy2cts
5,9642719
answered 11 hours ago
my2ctsmy2cts
5,9642719
answered 11 hours ago
my2ctsmy2cts
5,9642719
answered 11 hours ago
my2ctsmy2cts
5,9642719
5,9642719
add a comment |
add a comment |
triple7 is a new contributor. Be nice, and check out our Code of Conduct.
triple7 is a new contributor. Be nice, and check out our Code of Conduct.
triple7 is a new contributor. Be nice, and check out our Code of Conduct.
triple7 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
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– Kyle Kanos
11 hours ago
$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
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– jacob1729
8 hours ago
1
$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
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– G. Smith
5 hours ago
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No satellite orbits earth at an altitude of 705 of any commonly used unit--that's either so far out it's lost or it's deep under the surface. Kepler's third law is talking about the distance between the centers, not the distance between the surfaces.
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– Loren Pechtel
58 mins ago
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Aqua has a 95 minute orbit revolution and is at 705 km altitude above earth’s surface. I have taken into account earth’s radius...
$endgroup$
– triple7
42 mins ago