Posted by Jake on Wednesday, December 23, 2009 at 8:53pm.
i posted this earlier but am now stuck on getting through the algebra
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_moon/(384,403,000 m - r)^2 = m_Earth/r^2
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
( m_Earth(384,403,000 m)^2 - m_Earth(r^2) )/m_moon = r^2
(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 9:33pm
I 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 9:34pm
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 9:36pm
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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