How high do things bounce on the moon?

"Things bounce six times higher on the moon!"
James Corden, narrating Little Charley Bear

I have some serious problems with the quality of the physics shown on CBeebies. I'm happy to accept anthropomorphised animals, the idea that all of the presenters live in the CBeebies house, and I'll even entertain the idea that it's possible to turn it on without getting the theme tune to Balamory stuck in my head for days. But the way they treat rainbows is shocking. This isn't about rainbows, though, it's about the moon.

In an episode of Little Charley Bear, which Bill persists in liking despite my continual growling at it, the narrator (James Corden, presumably raking it in from licensing) asserts happily that 'things bounce six times higher on the moon.'

No, Mr Corden, no, they don't. If you drop a bouncy object from rest on the moon, assuming collisions with moon-rock are as elastic as they are on earth, it will bounce back to roughly the same height as it would on earth. Otherwise, you'd have a serious problem with the conservation of energy: you could drop a bouncy ball that (on Earth) bounces almost to the height you dropped it from, take it to the moon, and boom! it suddenly bounces six times higher. You could harness that extra energy on the way up, bounce it again, and get a perpetual motion machine.

The bounce would be slightly higher on the moon, because there's no air resistance to slow the ball down.

How high does a thing bounce?

Even if you project the ball downwards, the six times figure isn't quite right. By conservation of energy, the ball rebounds back to where you threw it from travelling at the same speed in the opposite direction -- and from there, rearranging $v^2 = u^2 + 2as$, you get $s = \frac{v^2 - u^2}{2a}$. You want to know where it stops, so $v^2=0$, and you're left with $s = -\frac{u^2}{2a}$.

Since the moon's gravitational pull is about a sixth of the earth's, you might think that Corden had it right, but I still say no: it bounces six times higher above your hand. That is to say, $S_m - S_0 \approx 6(S_e - S_0)$, or $S_m = 6S_e - 5S_0$. The bounce is six times higher, minus five times the height you dropped it from. Unless your thing would bounce many times higher on the earth than the height you flung it down from, the adjustment is significant enough that you need to account for it.

So, if you're reading this, Mr Corden, I think you need to get your pseudo-parenting skills sharpened up, pronto. Don't go filling Little Charley Bear's head with things that aren't true, or you will find, suddenly, that it is no longer compulsory to love him. And then where will your licensing rights be, hmm?

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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