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_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...
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
sorry
0 = (m_M r^2)/m_E + r^2 + 2(3.84403 m)r - (3.84403E9 m)^2
sorry 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
Algebra:
http://www.jiskha.com/display.cgi?id=1261622871
To continue solving the problem, we can simplify the equation further. Let's rewrite the equation as:
(m_Earth * (384,403,000 m)^2)/m_moon = (m_Earth (r^2) )/m_moon + r^2
First, let's simplify the left side of the equation:
(m_Earth * (384,403,000 m)^2)/m_moon = (m_Earth/Moon) * (384,403,000 m)^2
Next, let's simplify the right side of the equation:
(m_Earth (r^2))/m_moon + r^2 = (m_Earth + m_moon)/m_moon * r^2
Now, we have the equation:
(m_Earth/Moon) * (384,403,000 m)^2 = ((m_Earth + m_moon)/m_moon) * r^2
To isolate r^2, multiply both sides of the equation by m_moon:
(m_Earth/Moon) * (384,403,000 m)^2 * m_moon = (m_Earth + m_moon) * r^2
Divide both sides of the equation by (m_Earth + m_moon):
[(m_Earth/Moon) * (384,403,000 m)^2 * m_moon] / (m_Earth + m_moon) = r^2
Take the square root of both sides:
r = sqrt([(m_Earth/Moon) * (384,403,000 m)^2 * m_moon] / (m_Earth + m_moon))
Now, we can substitute the known values into the equation to calculate the distance from the Earth where the spacecraft will experience zero net force. Given:
Mass Moon (m_moon) = 7.35 E 22 kg
Mass Earth (m_Earth) = 5.98 E 24 kg
Distance Earth to Moon (r Earth to Moon) = 384,403,000 m
Plug in these values into the equation:
r = sqrt([(5.98 E 24 kg / 7.35 E 22 kg) * (384,403,000 m)^2 * 7.35 E 22 kg] / (5.98 E 24 kg + 7.35 E 22 kg))
After substituting the values, simplify the equation and calculate the square root to find the distance from the Earth where the spacecraft will experience zero net force.