- #1
Karol
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1. Homework Statement
Calculate the time the moon rotates once around the earth, and show that you have to add 4 hours to that since the moon & the Earth rotate around their center of mass. hint: use the reduced mass: [tex]\overline{m}[/tex]=[tex]\frac{m_{1}m_{2}}{m_{1}+m_{2}}[/tex]
mass of moon: m=0.0734E24[kg]
mass of earth: 81 times mass of moon: M=81m
radius of earth: Re=6.37E6[m]
distance centers earth-moon: 60Re
gravity constant: G=6.673E-11
(1) F=[tex]\frac{GMm}{r^{2}}[/tex]
(2) F=[tex]\frac{mV^{2}}{r}[/tex] (3) V=[tex]\frac{2r\pi}{T}[/tex]
xc.m.=[tex]\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}[/tex]
maybe: the momentum in the center of mass coordinate system: (4)[tex]\vec{p_{1}}[/tex]=m1[tex]\vec{V_{1}}[/tex]=[tex]\overline{m}[/tex][tex]\vec{V_{12}}[/tex], while V12 is the relative velocity between them, and [tex]\overline{m}[/tex]=[tex]\frac{m_{1}m_{2}}{m_{1}+m_{2}}[/tex] is the reduced mass.
the revolution time when the Earth is "stuck" in place and the moon rotates only: using equations: (1)=(2), (3): [tex]\Rightarrow[/tex] [tex]\frac{mV^{2}}{60R^{e}}[/tex]=[tex]\frac{81Gm^{2}}{(60R_{e})^{2}}[/tex] [tex]\Rightarrow[/tex] T=2,357,028[sec]=27.28[days], and the relative velocity is: V12=1018.8[m/sec].
the distance of the center of their mass from center of earth: xc.m.=[tex]\frac{60R_{e}m}{82m}[/tex]=4,660,976[m]
now, in the coordinate system of the moving center of gravity, i find the moon's tangential velocity, knowing that the relative velocity V12 is the same in all coordinate systems: according to equation (4) above: V1=[tex]\frac{\overline{m}V_{12}}{m_{1}}[/tex]=1006.3[m/sec]
now, the revolution time of the moon round the distance to center of mass (3): [tex]\frac{2\pi(60R_{e}-x_{c.m.})}{1006.3}[/tex]=2,357,187[sec]
this is only 159[sec] more, not 4 hours!
Calculate the time the moon rotates once around the earth, and show that you have to add 4 hours to that since the moon & the Earth rotate around their center of mass. hint: use the reduced mass: [tex]\overline{m}[/tex]=[tex]\frac{m_{1}m_{2}}{m_{1}+m_{2}}[/tex]
mass of moon: m=0.0734E24[kg]
mass of earth: 81 times mass of moon: M=81m
radius of earth: Re=6.37E6[m]
distance centers earth-moon: 60Re
gravity constant: G=6.673E-11
Homework Equations
(1) F=[tex]\frac{GMm}{r^{2}}[/tex]
(2) F=[tex]\frac{mV^{2}}{r}[/tex] (3) V=[tex]\frac{2r\pi}{T}[/tex]
xc.m.=[tex]\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}[/tex]
maybe: the momentum in the center of mass coordinate system: (4)[tex]\vec{p_{1}}[/tex]=m1[tex]\vec{V_{1}}[/tex]=[tex]\overline{m}[/tex][tex]\vec{V_{12}}[/tex], while V12 is the relative velocity between them, and [tex]\overline{m}[/tex]=[tex]\frac{m_{1}m_{2}}{m_{1}+m_{2}}[/tex] is the reduced mass.
The Attempt at a Solution
the revolution time when the Earth is "stuck" in place and the moon rotates only: using equations: (1)=(2), (3): [tex]\Rightarrow[/tex] [tex]\frac{mV^{2}}{60R^{e}}[/tex]=[tex]\frac{81Gm^{2}}{(60R_{e})^{2}}[/tex] [tex]\Rightarrow[/tex] T=2,357,028[sec]=27.28[days], and the relative velocity is: V12=1018.8[m/sec].
the distance of the center of their mass from center of earth: xc.m.=[tex]\frac{60R_{e}m}{82m}[/tex]=4,660,976[m]
now, in the coordinate system of the moving center of gravity, i find the moon's tangential velocity, knowing that the relative velocity V12 is the same in all coordinate systems: according to equation (4) above: V1=[tex]\frac{\overline{m}V_{12}}{m_{1}}[/tex]=1006.3[m/sec]
now, the revolution time of the moon round the distance to center of mass (3): [tex]\frac{2\pi(60R_{e}-x_{c.m.})}{1006.3}[/tex]=2,357,187[sec]
this is only 159[sec] more, not 4 hours!