- #1
paulcretu
- 8
- 0
Hi everybody,
I've registered to PF because after searching the web and this forum too, I couldn't find a clear answer to my dilemma. I've read this article but something seems wrong:
http://www.scientificamerican.com/article.cfm?id=half-mass-moon
What if the mass of the Moon would be lower, while Earth's mass is constant and also Moon's tangential velocity is constant? Would the Distance between the centers of Moon and Earth change? Would the (mean) Radius of Moon's orbit change?
According to the article above, "A less massive moon would also orbit closer to Earth than the real one". Well... how come and why? According to my calculations this is actually the other way.
First of all, I know that there are some factors I'm not taking into account (see below).
If we oversimplify, ignore the barycenter and consider Moon orbits the center of Earth then the answer is very simple. Orbit (distance between Moon's center and it's orbit rotation center) is the same as distance between centers of the two and the orbit (and distance) does not depend on Moon's mass, but only on it's velocity (and Earth's mass):
GmM/D^2 = mv^2/R and we approximate R=D, we raise by D^2 and simplify by m
GM=v^2 *D
D=GM/v^2 and it's the same no matter what is the mass of the Moon.
Now, considering the Moon orbits the barycenter, we have:
D = distance between centers,
R = radius of Moon's orbit / m = mass of Moon
r = radius of Earth, orbit / M = mass of Earth
r+R =D
R = D * M / (m + M) = D / (1 + m/M) ; the larger the mass difference between the two, the smaller m/M, and the more D is closer to R
we go back to GmM/D^2 = mv^2/R or
GM/D^2 = v^2/R
now we have:
GM/D^2 = v^2/[D/(1+m/M)]=(v^2)*(1+ m/M)/D
we multiply by D^2
GM=(v^2)*(1+m/M)*D
D = GM / [(v^2)*(1+m/M)] = GM / [((M+m)/M) * V^2] = GM^2 / [(m+M)v^2]
This would clearly show that if m becomes lower, D will be greater, in contradiction with the article mentioned above.
This doesn't take into account that:
- D is not constant (but I suppose its variations can be neglected)
- orbits are not perfect circles
- external influences and I don't know any other possible factors
BUT... I find it hard to believe the missing factors not taken into account could possibly affect the result so much that the conclusion would change.
So ... where did I go wrong? Please note that the article is signed or the information in it attributed to a University professor of physics and astronomy. Could the information be wrong?
Also please note:
R = GM^3/v^2(m+M)^2 so not only the distance will be larger, but also the barycenter moves towards Earth's center and the Moon's orbit Radius will be higher by a square factor.
Thanks in advance for any comments and replies.
I've registered to PF because after searching the web and this forum too, I couldn't find a clear answer to my dilemma. I've read this article but something seems wrong:
http://www.scientificamerican.com/article.cfm?id=half-mass-moon
What if the mass of the Moon would be lower, while Earth's mass is constant and also Moon's tangential velocity is constant? Would the Distance between the centers of Moon and Earth change? Would the (mean) Radius of Moon's orbit change?
According to the article above, "A less massive moon would also orbit closer to Earth than the real one". Well... how come and why? According to my calculations this is actually the other way.
First of all, I know that there are some factors I'm not taking into account (see below).
If we oversimplify, ignore the barycenter and consider Moon orbits the center of Earth then the answer is very simple. Orbit (distance between Moon's center and it's orbit rotation center) is the same as distance between centers of the two and the orbit (and distance) does not depend on Moon's mass, but only on it's velocity (and Earth's mass):
GmM/D^2 = mv^2/R and we approximate R=D, we raise by D^2 and simplify by m
GM=v^2 *D
D=GM/v^2 and it's the same no matter what is the mass of the Moon.
Now, considering the Moon orbits the barycenter, we have:
D = distance between centers,
R = radius of Moon's orbit / m = mass of Moon
r = radius of Earth, orbit / M = mass of Earth
r+R =D
R = D * M / (m + M) = D / (1 + m/M) ; the larger the mass difference between the two, the smaller m/M, and the more D is closer to R
we go back to GmM/D^2 = mv^2/R or
GM/D^2 = v^2/R
now we have:
GM/D^2 = v^2/[D/(1+m/M)]=(v^2)*(1+ m/M)/D
we multiply by D^2
GM=(v^2)*(1+m/M)*D
D = GM / [(v^2)*(1+m/M)] = GM / [((M+m)/M) * V^2] = GM^2 / [(m+M)v^2]
This would clearly show that if m becomes lower, D will be greater, in contradiction with the article mentioned above.
This doesn't take into account that:
- D is not constant (but I suppose its variations can be neglected)
- orbits are not perfect circles
- external influences and I don't know any other possible factors
BUT... I find it hard to believe the missing factors not taken into account could possibly affect the result so much that the conclusion would change.
So ... where did I go wrong? Please note that the article is signed or the information in it attributed to a University professor of physics and astronomy. Could the information be wrong?
Also please note:
R = GM^3/v^2(m+M)^2 so not only the distance will be larger, but also the barycenter moves towards Earth's center and the Moon's orbit Radius will be higher by a square factor.
Thanks in advance for any comments and replies.