A student says that of two pairs at the same separation, the heavier pair pulls harder. Is the student correct?
A heavier pair sat farther apart; a lighter pair sat closer together. Your two rules shrugged: they could not settle which force wins. Newton's law settles every such case, because it gives the force as a number.
You have seen that increasing either mass strengthens the force, and that distance weakens it by the inverse square: the 36 N force on the probe fell to 9 N at twice the distance. One equation ties the whole story together.
The two masses multiply together on top. Increase either one, and F grows.
The distance r is squared underneath. Double r, and F falls to one quarter: the inverse-square law you know.
r runs center to center: from the center of one object to the center of the other, in meters. r is never the surface-to-surface gap.
In F = G m₁m₂ / r², which distance is r?
G is a constant: the same multiplier in every gravity calculation, everywhere. G's unit is chosen so that F comes out in newtons. The value of G comes in the next lesson.
One of two moored ships takes on cargo, doubling its mass, the other ship and the distance unchanged. Use F = G m₁m₂ / r² to say what happens to the force between them.
Write down the equation:
F = G m₁m₂ / r²
Read off what changed:
m₁ doubles; m₂, r and G are unchanged.
The top of the fraction doubles.
The force F doubles.
In F = G m₁m₂ / r², how do the two masses enter?
The two ships then drift to twice their old separation. What does the law say now?
Write down the equation:
F = G m₁m₂ / r²
Read off what changed:
r doubles, so r² grows by 2² = 4.
The bottom of the fraction grows four times larger.
The force F falls to one quarter of its value.
In F = G m₁m₂ / r², how does the distance enter?
Which of the following is Newton's law of universal gravitation?
Which pairing of a symbol from F = G m₁m₂ / r² with its unit is correct?
A satellite orbits 3.0 × 10⁵ m above Earth's surface. Earth's radius is 6.4 × 10⁶ m. Enter the r for Newton's law, in meters.