- #1

GreenPrint

- 1,196

- 0

## 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

Last edited: