Calculating Earth's # of Rotations Around the Sun

[learningisfun]

I'm doing physics of uniform circular motion, while I'm working out this equation
It occurred to me , is it possible to find out how many rotations of Earth around the sun until it hits the sun. By the way I'm at a very NOOB level lol

but seriously?

*thinks of possible equation* :( not infromed well enough, maybe isolating r and increasing f. then finding the delta of r and ... i don't know

 

[mathman]

If there are no outside forces what makes you think that the Earth would ever hit the sun? The most likely end of the Earth scenario is the sun becoming a red giant, which might be big enough to swallow the earth.

 

[Chronos]

The Earth will actually recede from the sun over time.

 

[learningisfun]

well objects in uniform motion experience a force acting to accelerate the object towards the centere of curvature. Finding the Fg between the center of Earth and 1 kg on hte surface being 9.8N. Is their a cut of radius where the mass recedes from center?

 

[mgb_phys]

The Earth is in orbit around the sun - it won't hit it.
In fact, as Chronos said, because of tidal effects the Earth is slowly moving away from the sun.

 

[Division]

So it's more of a questions, how long is it until the Earth leaves the Sun's orbit, if we ever will.

 

[learningisfun]

wait

what is the graviational force between both spheres

m1=5.98x10^24 m2=7.35x10^22 r=3.84x10^8

Fg=Gm1m2/r^2

=6.67x10^-11(5.98x10^24)(7.35x10^22)/(3.84x10^8)^2
=3.41x10^30

the gravitational force between hte moon and the Earth is =3.41x10^30 (check my answer please to make sure I'm right, I'm not as handy with the calculator )

how would I find the distance or r where both forces councel eathother out?

distance between the Earth and moon at the center is 3.84x10^5 km = 3.84x10^8 m
earth/m1=5.98x10^24
moon/m2=7.35x10^22
G=6.67x10^-11would it be

give me a moment while I think
square root over

 

[learningisfun]

because Fg=Gm1m2/r^2= mv^2/r

thierfore

r=sqaurerootGm1m2/Fg

ok I got 2.93x10^14

this is murch greater then the distance between the moon and Earth at the center, I need to approach this differently. Somebody help me out >.>

hold on I have another fourmla in mind, to get r

r=Gm/v^2
in that case I would need to solve for v first

in this case
STATING VARAIBLES AGAIN ^.^
distance between the Earth and moon at the center is 3.84x10^5 km = 3.84x10^8 m
earth/m1=5.98x10^24
moon/m2=7.35x10^22
G=6.67x10^-11

v=square rootGm/r
=square root (6.67x10^-11)(5.98x10^24)/3.84x10^8 m
=1019m/s
the velocity of the moon is 1019m/s

plu this into r
r=square rootGm/r
=squareroot(6.67x10^-11)(5.98x10^24)/1019m/s
=6.25x10^5

DAMN >.< distance between moon and Earth is 3.82x10^5

 

[mgb_phys]

The point where the gravitational pull of the Earth and sun balance is called the L1 lagrange point.

 

[learningisfun]

I'm trying to find out the gravitational point of the moon and Earth it's one of my questions
-.- I did not realize that I forgot to include that in , ehm sorryFg=Gm1m2/r^2

=6.67x10^-11(5.98x10^24)(7.35x10^22)/(3.84x10^8)^2
=3.41x10^30

I'm not sure waht to do with Fg next

 

[learningisfun]

hmm, I have another equation in mind, take a moment to do it

 

[learningisfun]

This time I worked out the gravaitonal forces of the Earth and the moon. By dividing it I can gather a percentage of the difference in magnitude.

Fgearth/Fgmoon=%

G=6.67x10^-11
m1=5.98x10^24
m2=7.35x10^22
r1=6.38x10^6
r2=1.74x10^6

Fgearth=Gm1m/r1
=(6.67x10^-11)(5.98x10^24)(1)/6.38x10^6
=9.8N

Fgmoon=Gm2m/r2
=6.67x10^-11(7.35x10^22)(1)/1.74x10^6
=1.6N

Fgearth/Fgmoon=9.8B/1.6N=0.16%

Giving me a perecntage difference that I can use to find the distance and magnitude of both oposing forces.

Fg=Gm1m2/r^2

=6.67x10^-11(5.98x10^24)(7.35x10^22)/(3.84x10^8)^2
=3.41x10^30N

Fg gravity between Earth and the moon is 3.41x10^30N
Distance between spheres 3.84x10^8 m

0.84(3.84x10^8 m)=3.22x10^8
0.84(3.41x10^30N=3.86x10^30
Relative to earth

can somebody tell me if this is correct

 

[learningisfun]

apparently you could use quadratic equations, can someone explain how to do that

 

[HallsofIvy]

Use quadratic equations to do what?

 

[learningisfun]

Find the distance when the opposing forces of the Earth and the moon become neutral, relative to earth.
I did it in the above post, but I would like to do it differently using quadratic equations or substitution?

 

[Matterwave]

The only way I could imagine the Earth's orbit actually decaying would be from the loss of energy due to gravity waves. But this factor is so small that I think other factors are far larger by many orders of magnitude. Don't quote me on this though, I'm not a GR physicist.

 

[Nabeshin]

Find the distance when the opposing forces of the Earth and the moon become neutral, relative to earth.
I did it in the above post, but I would like to do it differently using quadratic equations or substitution?

Ok, first this has nothing to do with the orbital stability concern.

Your work is messy so I don't feel like looking through all of it but when you set up the equations yes you do get a quadratic because you're looking at the forces as scalars rather than vectors (In reality, there is only one point at which the forces cancel out but in the analysis we typically just set the magnitudes of the forces equal, which happens in two different positions). The point you're looking for will always be the smaller number of the two.

 

[learningisfun]

ok

The neutral point between the moon and the earth

m(moon)7.35E22
m(earth)5.98E24
R(between Earth and moon)3.84E5km=3.84E8m
G6.67E-11v


I used qudratics this time , now I'm stuck and will tackle it again after i Get some help
^.^


GMmM/r^2=GMeM/(L-r)^2
Mm/r^2=Me/(L-r)^2
r^2Me=Mm(L-r)^2
r^2Me=5.98E24(3.84E8m-r)^2
expand and simplfy
r^2Me=(5.98E24r^2+4.6E33r+8.97E41)

I'm not sure what to do next, I think I'm suppose to expand and simply the left side, then use algebra to substitute it in

Can someone clarify this for me,I'm trying

 

[Chronos]

the moon recedes from the Earth at nearly 4 cm per year. this effect has been confirmed by the lunar lasar ranging experiment. see http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/tides.html
and
http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_Experiment

 

[learningisfun]

hmm, ok that is interesting to read. This does not answer my question?

actaully hold on, I"m working on it