You are investigating Newtons Law of Gravity (
𝐹
𝐺
=
𝐺
𝑚
𝑀
𝑟
2
). Lets assume you measured the attractive force FG for two known masses m and M for different distances r. You would like to find an experimental value for
𝐺
, the gravitational constant.

You have several options to plot the gravitational force FG (y-axis) and distance r (x-axis). Which of these plots will most easily provide an experimental value for
𝐺
?

Question

You are investigating Newtons Law of Gravity (
𝐹
𝐺
=
𝐺
𝑚
𝑀
𝑟
2
). Lets assume you measured the attractive force FG for two known masses m and M for different distances r. You would like to find an experimental value for 
𝐺
, the gravitational constant.

You have several options to plot the gravitational force FG (y-axis) and distance r (x-axis). Which of these plots will most easily provide an experimental value for 
𝐺
?

BlackTom AI · Accepted Answer

FG = G m M / r^2 shows FG is directly proportional to 1/r^2, so plotting FG versus 1/r^2 will give a straight line whose slope contains G.

1) Relevant concepts
- Newton’s law of gravitation: FG = G m M / r^2.
- Linearization: to extract constants experimentally, rearrange into y = slope · x form.
- Slope and intercept of a linear fit give measurable constants (here slope = G m M).

2) Step-by-step reasoning
Start from FG = G m M / r^2. Define x = 1/r^2 and y = FG. Then y = (G m M) x. This is a linear equation through the origin (intercept zero), so a plot of y versus x will be a straight line with slope S = G m M.

3) Algebra / intermediate steps
FG = G m M / r^2
Let x = 1/r^2, y = FG
y = G m M · x
Slope S = G m M
Therefore G = S / (m M).

4) Why this is correct
Because plotting FG vs 1/r^2 converts the inverse-square dependence into a direct proportionality. A linear fit yields slope S directly equal to G m M, so calculating G requires only dividing the measured slope by the known product m·M. Noise may introduce a small intercept, but the slope still gives the best estimate of G.

5) Why other plots are less suitable
- FG vs r: FG ∝ 1/r^2 produces a curved (nonlinear) plot, not convenient for linear fitting.
- FG vs r^2: gives FG = (G m M)/(r^2) = (G m M)/x if x = r^2, an inverse relation (hyperbola), not linear.
- FG vs 1/r: gives FG ∝ 1/r, still not linear with exponent −2.
- Log-log (log FG vs log r) would give a straight line with slope −2 and intercept log(G m M) — this is a valid alternative, but requires extra transformation and extraction of G from the intercept, so FG vs 1/r^2 is the simplest direct method.

6) Conclusion
Plotting FG versus 1/r^2 yields a straight line with slope = G m M, so G = slope/(m M). This confirms the correct choice: FG vs r^(−2).

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