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Distance fallen

Gravity and orbits · level 5, lesson 1 of 6

A question to start

A climber on Mars drops a carabiner from a cliff, and it takes 4 seconds to hit the ground. How high is the cliff? The carabiner did not move at one speed: it started at zero and kept speeding up. How do you get a distance from that?

The idea

A dropped thing starts at rest and gains speed steadily: \(g\) metres per second, every second. After \(t\) seconds, its speed is

\[v = g t.\]

Because the speed grows steadily, its average over the fall is halfway between the start (0) and the end (\(gt\)):

\[\text{average speed} = \frac{0 + gt}{2} = \tfrac12 g t.\]

Distance is average speed times time, so

\[s = \tfrac12 g t^2.\]
gttdistance = ½ · t · gtspeedtime

The same answer shows as a picture: on a speed-time graph, distance is the area under the line. Here it is a triangle with base \(t\) and height \(gt\), area \(\tfrac12 \cdot t \cdot gt\).

Time \(t\) 1 s 2 s 3 s 4 s 5 s
Fallen on Mars (\(g\) = 3.71) 1.86 m 7.42 m 16.7 m 29.7 m 46.4 m
Fallen on Earth (\(g\) = 9.81) 4.91 m 19.6 m 44.1 m 78.5 m 123 m

Time for a given drop. Turn the formula round:

\[t = \sqrt{\frac{2s}{g}}.\]

For the same drop on two planets, the times are in the ratio \(\sqrt{g_E/g_M} = \sqrt{9.81/3.71} \approx 1.63\). Every fall on Mars takes about 1.6 times as long as the same fall on Earth: not 2.6 times, because of the square root.

These formulas ignore air. On Mars the air is so thin that they work well for heavy things over quite large drops.

Worked examples

1. The cliff. \(s = \tfrac12 \times 3.71 \times 4^2 = 1.855 \times 16 = 29.7\) m.

2. A dropped tool from a habitat roof, 6 m high. Mars: \(t = \sqrt{2 \times 6 / 3.71} = \sqrt{3.23} = 1.80\) s. Earth: \(t = \sqrt{12 / 9.81} = \sqrt{1.22} = 1.11\) s. Ratio \(1.80 / 1.11 = 1.62\). ✓

3. Finding \(g\) from a drop. On an unknown world, a stone falls 10 m in 2.0 s. Then \(10 = \tfrac12 g \times 4\), so \(g = 5\) m/s².

Try it

  1. How far does a stone fall in 3 s on Mars?
  2. How long does a stone take to fall 50 m on Earth? On Mars?
  3. A stone falls 20 m in 3.3 s on some moon. Find \(g\) there.
  4. Show from the formula that doubling the time makes the distance four times as large.
  5. In the 5th second alone, how far does a stone fall on Mars?
Answers
  1. \(1.855 \times 9 = 16.7\) m.
  2. Earth: \(\sqrt{100/9.81} = \sqrt{10.19} = 3.19\) s. Mars: \(\sqrt{100/3.71} = \sqrt{26.95} = 5.19\) s.
  3. \(g = 2s/t^2 = 40 / 10.89 \approx 3.7\) m/s². (That moon has about the gravity of Mars.)
  4. \(\tfrac12 g (2t)^2 = \tfrac12 g \cdot 4t^2 = 4 \times \tfrac12 g t^2\).
  5. \(46.4 - 29.7 = 16.7\) m.

Common mistake

Using the final speed for the whole fall. A stone that lands at 14.8 m/s after 4 s on Mars did not fall \(14.8 \times 4 = 59\) m. It was slower earlier; its average speed was half that, so it fell 29.7 m.

Where it leads