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April 17, 2015

Posted by **Jake** on Wednesday, December 23, 2009 at 8:53pm.

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. 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 further

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

add (m_Earth (r^2) )/m_moon to both sides

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

I don't know were to go from here...

- physics -
**Jake**, Wednesday, December 23, 2009 at 9:33pmI believed this is correct after noticing my mistake

Ok I'm here what do I do now

0 = (m_M r^2)/m_E + r^2 + 2(3.84403 m)r - (3.84403E9 m)2

- physics -
**Jake**, Wednesday, December 23, 2009 at 9:34pmsorry

0 = (m_M r^2)/m_E + r^2 + 2(3.84403 m)r - (3.84403E9 m)^2

- physics -
**Jake**, Wednesday, December 23, 2009 at 9:36pmsorry again

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

sorry about that

- See other post -
**MathMate**, Thursday, December 24, 2009 at 8:06am

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