Gravitation and Newton's Synthesis

[GreenPrint]

Homework Statement

13) (II) At what distance from the Earth will a spacecraft on the way to the Moon experience zero net force due to these two bodies becasue the Earth and Moon pull with equal and opposite forces?

Homework Equations

NET F = ma
G = 6.67 E-11 (Nm^2)/kg^2
Fg = (GmM)/r^2
Mass Moon = 7.35 E 22 kg
Mass Earth = 5.98 E 24 Kg
r Earth to Moon = 384,403,000 m

The Attempt at a Solution

Apply Newton's s second law in the radial direction

NET F = m_craft( a_radial) = Fg moon = Fg Earth = 0
= (G m_craft m_moon)/(384,403,000 m - r)^2 = (G m_craft m_Earth)/r^2

m_craft cancels
G cancels

m_moon/(384,403,000 m - r)^2 = m_Earth/r^2
simplify

m_moon/((384,403,000 m)^2- r^2) = m_Earth/r^2

raise both sides to negative one power

((384,403,000 m)^2- r^2)/m_moon = r^2/m_Earth

multiply both sides by m_Earth

m_Earth( (384,403,000 m)^2 - r^2 )/m_moon = r^2

simplify

( m_Earth(384,403,000 m)^2 - m_Earth(r^2) )/m_moon = r^2

simplify

( m_Earth(384,403,00 m)^2 - 2(m_Earth)(384,403,000 m)m_Earth(r^2) - m_Earth(r^2) )/m_moon = r^2

simplify further

( m_Earth(384,403,00 m)^2 - 2(m_Earth)^2(384,403,000 m)(r^2) - m_Earth(r^2) )/m_moon = r^2

multiply both sides by 1/m_Earth and m_moon to clean up

(384,403,00 m)^2 - 2(m_Earth)(384,403,000 m)(r^2) - (r^2) = (m_moon (r^2))/m_Earth

 

[denverdoc]

Do you know how to use the quadratic equation?

 

[GreenPrint]

yes i do

 

[cepheid]

m_moon/(384,403,000 m - r)^2 = m_Earth/r^2
simplify

m_moon/((384,403,000 m)^2- r^2) = m_Earth/r^2

NO.

Recall that: (a - b)² ≠ (a² - b²)

By definition:

(a - b)² = (a - b)(a - b)

= aa - ab - ba + bb

= a² - 2ab + b²

Another tip:

Don't bother plugging in numbers and carrying them through all of those intermediate steps. It just clutters things up! Work through the entire problem algebraically (i.e. in symbols) until the end, when you have solved for an expression for r, the distance where the forces are equal. The nice thing is that this expression will be general. It will give the distance between any two masses m1 and m2 (with a given separation R) at which the gravitational forces cancel. Then you can plug in specific numbers for the case of Earth and its moon.

 

[denverdoc]

Good, because that is what I think will finish the problem: but before going there check your work basically you have A*(k-r)^2=B*r^2 Seems you might have dropped some terms: the left side would be A(k^2-2akr+r^2).

Edit: just as Cepheid points out, and the tip should be taken to heart: keep the numbers out of it until the end--lots of wasted writing, and makes it MUCH harder to follow.

 

[GreenPrint]

Ok I'm here what do I do now

0 = (m_M r^2)/m_E + r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2

sorry about that

 

[GreenPrint]

There that should be correct sorry

 

[GreenPrint]

I was wrong

0 = (m_M r^2)/m_E - r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2

i believe this is correct

 

[denverdoc]

Assuming you have done the math right, which I'll check in a minute--

you have r^2(m/M-1) + 2d*r - c^2. Use the quad, and see if your answer makes sense--you can check the work yourself by plugging in the distance in your original formula--forces should balance.

That looks better, I thought you had flipped a sign, remember it is so much easier keeping the numbers out of it!

 

[GreenPrint]

I think I did it right

 

[denverdoc]

excellent--again your answer can be put back in the original eqn as a check. Really do give some effort at keeping the numbers out of it until you're ready to use your calculator. Errors are so much easier to spot and your prof will be appreciative as well!

 

[GreenPrint]

were do i go from here

 

[denverdoc]

were do i go from here

To the next problem, right? I'm assuming you used the quadratic formula to solve for r.

Recall (-b +/- sqrt(b^2-4ac))/2a Now the real fun begins.

 

[GreenPrint]

I don't know which one is a, b or c here becasue there are four terms...

 

[GreenPrint]

please! I'm almost done

 

[denverdoc]

Collect the terms around r^2 This is a. The term in front of r is b and the pure number is c.

 

[GreenPrint]

and how do I do that

0 = (m_M r^2)/m_E - r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2

there are two terms around r^-2

m_M r^2/m_E and
- r^2
all contain an r
how do i just simply collect this

 

[denverdoc]

Notice what i did way back when -- I reposted your work as r^2(m/M-1) where m is the little mass, and M is the big mass. The quantity (m/M-1) is A. BTW that's ((m/M) - 1)

 

[GreenPrint]

how did you go from this

0 = (m_M r^2)/m_E - r^2 + 2(3.84403 E 8 m)r - (3.84403 E8 m)^2

to this

((m/M) - 1)

for the first term and getting rid of the second by combining it with the first to get

((m/M) - 1)


i don't see were that comes from

thanks

 

[ideasrule]

(m_M r^2)/m_E - r^2
=r^2 (m_M/m_E-1)

 

[GreenPrint]

oh wow i feel dum...

thanks

 

[GreenPrint]

but why is this a true statement?

(m_M r^2)/m_E - r^2
=r^2 (m_M/m_E-1)

 

[denverdoc]

I just moved the r^2 in front and represented the two terms as a product of the common factor r^2 and m/M and -1

Multiply r^2(m/M-1) you will be doing exactly the oppsite of what i did.

 

[GreenPrint]

oh gee I feel stupid :O

 

[denverdoc]

Oh don't feel bad, i personally think your notation while elegant is just getting in your way. And maybe the algebra is rusty.

I like to keep it simple. I do have a suggestion--quadratics like this are real easy to bung up unless you have a programmable calculator. If you don't i found this applet:

http://mste.illinois.edu/users/exner/ncsa/quad/

That way you won't be wasting hours potentially. Just evaluate A. B, C and plug em in.

 

[ideasrule]

Now that I've actually read the question, there's an easier (MUCH easier) way of doing it. Assume that the spacecraft is at d1 from Earth and d2 from the Moon. Then you can figure out d1/d2 by doing GMe/d1^2 = GMm/d2^2. Since you now know the values of d1/d2 and d1+d2, solving for d1 and d2 is easy.

 

[denverdoc]

Nice job as essentially (d1/d2)=sqrt(m/M) so solving simultaneously gives rise to

d1+d1/sqrt(m/M)=total distance.