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Falling, on Earth and on Mars

Change and motion · level 5, lesson 5 of 6

A question to start

Drop a stone down a deep well and count the seconds until you hear the splash. If it takes 2.5 seconds, how deep is the well? You cannot just multiply a speed by the time, because the stone has no single speed: it starts at zero and keeps getting faster.

Now take the same stone to Mars and drop it down a well of the same depth. Will it take a little longer, or much longer? This lesson finds a short rule for falling, explains why it is true, and uses it on two planets.

The idea

What falling looks like

Imagine a camera that takes a picture of a falling stone once every second, all on the same photo. This is called a strobe picture. Here are a stone on Earth and a stone on Mars, drawn to the same scale.

1 s2 s3 s4 s: 78.4 m 2 s3 s4 s: 29.7 m EarthMars

On both planets the gaps between the dots grow. The stone covers more ground each second than the second before, because it keeps speeding up. On Mars everything happens more gently.

The rule for Earth

Measure the Earth stone and you find (ignoring air resistance):

Time \(t\) (s) 0 1 2 3 4 5
Distance fallen (m) 0 4.9 19.6 44.1 78.4 122.5
Fallen in that second (m) 4.9 14.7 24.5 34.3 44.1
Increase on the second before (m) 9.8 9.8 9.8 9.8

Divide each distance by 4.9: you get 0, 1, 4, 9, 16, 25, the square numbers. So

\[d = 4.9\,t^2 \text{ metres (on Earth)}.\]

Look at the last row. The first row of differences (4.9, 14.7, 24.5, ...) is not constant: the change is not steady, so the graph is a curve. But the differences of the differences, called the second differences, are all 9.8. That pattern, constant second differences, is the fingerprint of a rule with a \(t^2\) in it (you meet it again in Patterns and Algebra).

Why the rule is true

Here is a reason, not just a pattern. From lesson 4: a falling stone gains 9.8 m/s of speed every second. So, starting from rest, its speed after \(t\) seconds is

\[\text{speed} = 9.8\,t \text{ m/s}.\]

The speed rises steadily, from 0 at the start to \(9.8t\) at the end. When something changes steadily, its average over the time is exactly halfway between the start and the end. (Think of the speeds second by second: every speed in the early half is matched by one in the late half that is just as far above the middle.) So the average speed is

\[\frac{0 + 9.8t}{2} = 4.9\,t \text{ m/s}.\]

And distance is average speed times time:

\[d = 4.9\,t \times t = 4.9\,t^2.\]

The 4.9 is half of 9.8. That is where it comes from: the stone's average speed is half its final speed. In lesson 6 you will see the same argument as a picture, a triangle under a speed-time graph.

Written with the letter \(g\) for the acceleration due to gravity, the rule for anything falling from rest is

\[d = \tfrac{1}{2}\,g\,t^2.\]

The rule for Mars

Mars has about one ninth of the Earth's mass but just over half its width, and the result is a pull at the surface of about \(g = 3.71\) m/s². That is \(3.71 \div 9.81 \approx 0.38\), roughly 38% of Earth's. Half of 3.71 is 1.855, so

\[d \approx 1.86\,t^2 \text{ metres (on Mars)}.\]
Time \(t\) (s) 0 1 2 3 4 5
Distance fallen on Mars (m) 0 1.86 7.44 16.74 29.76 46.5
Fallen in that second (m) 1.86 5.58 9.30 13.02 16.74
Second differences (m) 3.72 3.72 3.72 3.72

The second differences are 3.72, twice 1.86. (They would be exactly 3.71 if we had kept 1.855; the 0.01 comes from rounding.)

Earth: 4.9t²Mars: 1.86t² 01234 time (seconds) 020406080 m

Same time, same drop: comparing the planets

Two neat facts fall out of the rules.

  • In the same time, a Mars stone falls \(1.86 \div 4.9 \approx 0.38\) as far as an Earth stone: the same 38%, because both rules have the same \(t^2\) and only the number in front differs.
  • For the same drop, the Mars stone takes longer, but not 1/0.38 times as long. If \(4.9\,t_E^2 = 1.86\,t_M^2\), then \(t_M^2 \div t_E^2 = 4.9 \div 1.86 \approx 2.63\), so \(t_M \div t_E \approx \sqrt{2.63} \approx 1.62\). Every fall on Mars takes about 1.6 times as long as the same fall on Earth.

The square root appears because time is squared in the rule. Gravity 2.6 times weaker does not mean falls 2.6 times slower.

Heavy and light

Does a heavy stone fall faster than a light one? Without air, no: everything near the surface gains speed at the same rate \(g\). The story goes that Galileo dropped two balls of different weights from the Leaning Tower of Pisa to show this; historians doubt he really did, but he did study falling carefully with balls rolling down ramps. In 1971 the astronaut David Scott of Apollo 15 dropped a hammer and a falcon feather side by side on the Moon, where there is no air, and they hit the ground together. On Earth the feather drifts because of the air, not because gravity treats it differently. Mars has a thin atmosphere, so a feather there would still drift, just less.

Worked examples

1. The well. A stone takes 2.5 s to reach the water. How deep is the well (ignoring the time the sound takes to come back up)?

\(d = 4.9 \times 2.5^2 = 4.9 \times 6.25 = 30.625\), about 31 m. (Sound takes about 0.09 s to climb 31 m, so the true fall time is a little under 2.5 s and the well is slightly shallower, about 29 m.)

2. A cliff on two planets. A rock falls from a 45 m cliff. How long does it take on Earth and on Mars, and how fast is it going at the bottom?

  • Earth: \(4.9t^2 = 45\), so \(t^2 = 45 \div 4.9 \approx 9.18\) and \(t \approx 3.03\) s. Speed \(= 9.8 \times 3.03 \approx 29.7\) m/s.
  • Mars: \(1.855t^2 = 45\), so \(t^2 \approx 24.26\) and \(t \approx 4.93\) s. Speed \(= 3.71 \times 4.93 \approx 18.3\) m/s.

Check the 1.6 rule: \(3.03 \times 1.62 \approx 4.91\). ✓ (close; the small gap is rounding).

3. The ruler test of reaction time. A friend holds a ruler; you hold your fingers at its zero mark. They drop it without warning and you catch it after it has fallen 20 cm. What is your reaction time?

\(4.9t^2 = 0.20\), so \(t^2 = 0.20 \div 4.9 \approx 0.0408\) and \(t \approx 0.20\) s. A fifth of a second.

4. The fifth second. How far does an Earth stone fall during the fifth second alone (from \(t = 4\) to \(t = 5\))?

\(122.5 - 78.4 = 44.1\) m. Using the differences: each second's distance is 9.8 m more than the last, and the fourth second was 34.3 m, so the fifth is \(34.3 + 9.8 = 44.1\) m. ✓

Try it

  1. How far does a stone fall in 6 s on Earth? On Mars?
  2. How long does a stone take to fall 78.4 m on Earth? On Mars (to 1 decimal place)?
  3. In the ruler test, a friend catches the ruler after 15 cm. Find their reaction time (to 2 decimal places).
  4. How fast is a stone falling on Mars after 2.5 s?
  5. On Earth, how much further does a stone fall in the 8th second than in the 7th? Explain without working out either distance.
  6. Explain in your own words why the number in \(d = 4.9t^2\) is half of 9.8.
  7. A ball rolls down a ramp from rest and covers 0.3 m in the first second. Assuming steady acceleration, how far has it gone after 2 s, 3 s and 4 s?
  8. A rover's camera mast is 2 m high. If a small part came loose and fell from the top, how long would it take to reach the ground on Mars? How much shorter would the same fall be on Earth?
Answers
  1. Earth: \(4.9 \times 36 = 176.4\) m. Mars: \(1.86 \times 36 = 66.96\) m, about 67 m.
  2. Earth: \(t^2 = 78.4 \div 4.9 = 16\), so \(t = 4\) s. Mars: \(t^2 = 78.4 \div 1.86 \approx 42.15\), so \(t \approx 6.5\) s.
  3. \(t^2 = 0.15 \div 4.9 \approx 0.0306\), so \(t \approx 0.17\) s.
  4. \(3.71 \times 2.5 \approx 9.3\) m/s.
  5. 9.8 m. The distance in each second is 9.8 m more than in the second before (the second differences are all 9.8), because the stone is 9.8 m/s faster on average.
  6. The speed rises steadily from 0 to \(9.8t\), so the average speed is half the final speed, \(4.9t\). Distance is average speed times time, \(4.9t \times t\).
  7. Distance grows like \(t^2\): \(0.3 \times 4 = 1.2\) m, \(0.3 \times 9 = 2.7\) m, \(0.3 \times 16 = 4.8\) m.
  8. Mars: \(t^2 = 2 \div 1.855 \approx 1.078\), so \(t \approx 1.04\) s. Earth: \(t^2 = 2 \div 4.9 \approx 0.408\), so \(t \approx 0.64\) s. About 0.4 s shorter on Earth (and \(1.04 \div 0.64 \approx 1.6\), the same ratio as always).

Common mistake

Using distance = speed × time with the final speed. A stone that hits the water at 24.5 m/s after 2.5 s did not fall \(24.5 \times 2.5 = 61\) m. It was slower earlier. With steady acceleration from rest, use the average speed, which is half the final speed: \(12.25 \times 2.5 \approx 30.6\) m.

Scaling time like gravity. Mars gravity is about 0.38 of Earth's, but falls do not take \(1 \div 0.38 \approx 2.6\) times as long. Because time is squared in the rule, they take \(\sqrt{2.6} \approx 1.6\) times as long.

Where it leads