ASTR220 Collision In Space – Handout #4
Working With Proportionalities and Ratios
Often a mathematical problem is made easier by taking advantage of proportionalities or considering
ratios of quantities. Take, for example, Newton’s Law of Gravity:
F = G M1 M2 / R2.
This equation gives the force felt between two particles of mass M1 and M2, separated by a distance
R. In the metric system, the force would be measured in Newtons (N), the masses in kilograms (kg),
and the distance in meters (m). The G in the equation is the so-called “universal constant of
gravitation” needed to make the units work out. In the metric system, G = 6.672  10-11 m3 kg-1 s-2, a
very unwieldy quantity indeed.
If you are only interested in how, say, the force changes when you use objects of different mass, or
give them a different separation, there is no need to worry about G. Instead, just note that the force
is proportional to the product of the masses and the inverse square of the separation, i.e. F  M1 M2 /
R2. This means if you double M2, then F will be twice as large. If you put the masses twice as far
apart, then F will be 1/(22) = ¼ (one quarter) as large.
PRACTICE EXERCISE: How does F change if both M1 and M2 are 10 times larger, but
separated by 10 times the distance?
If you would like to know how, say, the gravitational force between the Earth and the Sun compares
to the force between Jupiter and its moon Io, then you can compute ratios by dividing the expression
for Jupiter and Io by the expression for the Earth and the Sun:
(FJupiter-Io/FEarth-Sun) = (MJupiter/MSun) (MIo/MEarth) (REarth-Sun)2 / (RJupiter-Io)2.
Notice that G drops out when you take the ratio (for this reason, you can also take the ratio of
proportionalities without knowing the exact underlying equations, i.e. without knowing the value of
G in this case). All you have to do now is make sure you use consistent units for mass and distance.
Jupiter is 1/1000th the mass of the Sun and (looking up in a textbook), Io is 0.015 the mass of Earth.
The Earth-Sun distance is 1 AU (by definition), and Io is 0.0028 AU from the center of Jupiter.
Plugging these numbers into the equation gives (FJupiter-Io/FEarth-Sun) = 10-10, in other words, the force
between Jupiter and Io is much weaker than the force between the Earth and the Sun. This makes
sense since our planet is very massive compared to Io, and very far from the Sun compared to Io’s
distance from Jupiter, so the Sun must exert a much greater force to keep us in orbit.
PRACTICE EXERCISE: How does the force ratio change if Io is 100 times more massive than
it actually is, and only 1/10th of its present distance from Jupiter?
By casting your equations as proportionalities or ratios, you can save a lot of work, and you can
reduce errors arising from unit conversions. It also allows you to concentrate on the physics that the
equations are describing, rather than the arithmetic.