< Gravity Examples

The Astronaut between Earth and Moon

Here is a problem which envisions an unfortunate astronaut stranded exactly half way between the earth and moon. What is the combined force of the earth and moon on it?
The data we need are When doing calculations like this is is usually a good idea to
  1. Keep units with the quantities used and check that the result has the correct units.
  2. Gather the powers of ten and calculate the exponent of the power of ten separately from the mantissas (the numbers in front of the powers of ten.)
Here is the calculation for the force due to earth:
$$F_g=G \frac{m_a m_{earth}}{r_{12}^2}$$ $$= \left( 0.667 \times 10^{-10} \frac{\hbox{N m}^2}{\hbox{kg}^2}\right)\frac{(100 \hbox{kg} ) (6\times 10^{24} \hbox{kg} )}{\left(\frac{1}{2} \times 3.8\times 10^8 \hbox{m} \right)^2}$$ $$= \frac{(0.667)(100)(6)}{1.9^2} \times 10^{-10}10^{24}(10^{-8})^2 \frac{[\hbox{N m}^2/\hbox{kg}^2] \hbox{[kg][kg]}}{\hbox{m}^2}$$ $$=110 \times 10^{-10+24-16} \hbox{N} = 1.1 \:\hbox{N}$$
Thus our 100-kg astronaut only feels a force of 1 N from the earth!

What is the force of the moon at the halfway point? To figure this out I would like to avoid that messy calculation again. Since all the values are the same except the mass of the moon replaces the mass of the earth Just use the relation
$$F_{moon} = F_{earth} \frac{m_{moon}}{m_{earth}}$$
= 1.1 N (7.2×1022)/(6×1024)

= 0.013 N

The total force must be obtained by adding these as vectors. They are acting in exactly opposite directions. The force from the moon is smaller that the lowest significant figure we calculated for the force from the earth so

Ftotal = 1.1 N - 0.013 N = 1.1 N ( about )


(The effect of the moon is insignificant to 1 percent.)

Problem

How far from the earth and from the moon must the astronaut be in order for the net force to be zero? (Assumme the astronaut is directly between the earth and moon.)