^{note 1}).

And yet, here he was, last Sunday night on

*Coast to Coast AM*, going one better. Just like last time, he was only summarizing what is on his web site so you can get the whole thing without having to slog through two hours of audio.

He seeks to explain why so many Mars landers have crashed instead of soft-landing as intended (10 out of 18, according to him, and for all I know it may be the correct figure). Here are the steps in what he laughingly calls his "logic":

- The gravitational attraction a planet has for an object in its vicinity is only partly described by the Newtonian force G.m1.m2/d
^{2}. - There's an additional term to consider, the centrifugal force generated by the planet's rotation.
- This force acts in opposition to the Newtonian force.
- Since Mars' rotation rate is 2.5% slower than that of Earth (actually the firgure is 2.8%) it generates less centrifugal force, and therefore more
*effective*gravity, in its vicinity than Earth. - An incoming lander is subject to the sum of the forces of gravity and centrifugal.
- Therefore a terminal flight profile calculated to be correct in Earth conditions fails on Mars.

**Is he right? No, of course he isn't**

Well, first off let me say that centrifugal force acting in opposition to the force of gravity on the surface of a planet is, indeed, a reality. "It's a thing," in the slang of today. This force can readily be calculated; it is

**-(mv**

^{2 }**cos L)**

**/r**

where

**m**is the mass of an object on the surface of a rotating planet

**v**is the linear velocity of the planetary surface at the equator (465 m/sec for Earth)

**r**is the radius of the planet (6.378 x 10

^{6}m for Earth)

**L**is the latitude where the force is measured

For Earth, the (

**v**term works out as 0.034 m/sec

^{2}cos L)/r^{2}at the equator where cos L evaluates to 1. A body, such as a fat woman, of mass 100kg weighs 340 grams less at the equator than at the poles, where cos L, and the centrifugal force, are both zero.

^{note 2}

Statements 1 & 2 are therefore in general correct when considering an object on a planet's surface. Statement 3 is also correct—it's perfectly possible to imagine a planet that rotates so rapidly that anything not tied down at its equator would be flung off into space. We would say that centrifugal force exceeds the force of gravity, in such a case.

^{note 3}

Statement 4 is a problem although Cotterell is basically correct in writing that centrifugal force is less on Mars. It has as much to do with the smaller size of the planet as with its rotation rate. However, that small difference is swamped by the fact that Newtonian gravity is very much less. Here are the figures (at the equator in both cases):

Earth, acceleration due to gravity: 9.863 m/sec

^{2}

Earth, acceleration due to centrifugal force: -0.034 m/sec

^{2}

Net acceleration: 9.829 m/sec

^{2}

Mars, acceleration due to gravity: 3.721 m/sec

^{2}

Mars, acceleration due to centrifugal force: -0.0171 m/sec

^{2}

Net acceleration: 3.704 m/sec

^{2}

It's in writing Statement 5 that Cotterell has gone completely haywire. He writes

*"Newton failed to recognize, in his equation, that a falling body is also under the influence of 'centrifugal force' caused by the spinning of the Earth on its axis."*He's taken the purely local and surface-based phenomenon of centrifugal force, and made it a property of the planet as a whole, extending beyond the surface into the region where incoming landers start feeling the effect of a planet's gravity. This is as preposterous as Cotterell's prior comments about gravity, and shows complete lack of understanding of physics. Of course a spacecraft having no physical contact with a planet cannot possibly be influenced by rotation of the planet. Neither can a falling apple, come to that, so Newton's equation describes that event accurately.

^{note 4}

Statement 6 suggests that engineers devising flight profiles for soft landings simply don't know about this, and therefore miscalculate. Last Sunday night, even that old softie George Noory demurred in the gentlest possible way. He reminded Cotterell that the landing of MSL and its rover

*Curiosity*in Gale crater six years ago (almost to the day, actually) had been a brilliant success and by no means a miscalculation. Cotterell mumbled something about engineers having learnt that when they completed their calculations they should "add a little bit, just for luck."

Cotterell is kind-of entertaining I suppose, with his bluff manner and his soft Lancashire accent, but he should be permanently banned from the fields of physics and mathematics lest he do even more damage to them.

=================/ \=====================

[1] "Matilda told such dreadful lies"

[2] In fact, there's another phenomenon that affects the acceleration due to gravity on the planet's surface. The equatorial radius is 6378 km but the polar radius is only 6357 km. Since the fat woman is 21 km further away from the center of the planet when she's on the equator than when she's at the poles, gravity has less of a hold on her. The effect amounts to 0.668%.

[3] I'm going to be in trouble with the purists for even talking in terms of "centrifugal force." To them, this force is merely the "equal and opposite" reaction to a centri

*petal*force. They would prefer to say "The force of gravity is inadequate to provide the centripetal force needed to keep objects attached to the planet." See the difference? But Cotterell uses centrifugal, and it's intuitive, so I'm going to stick with it.

[4] Note that as long as the apple is attached to the tree, it is pulled upwards by the small amount attributable to centrifugal force. As soon as it detaches, however, that small upward force vanishes.

## 8 comments:

Interesting, thought provoking article, and thanks for the formula.

But...

Sorry, expat, maybe it's not just Cotterell who doesn't quite get this.

Yes, a directly (perpendicularly to the landing point) approaching Lander, outside the planet's atmosphere cannot be affected by the 'centrifugal' force you discuss. (Note 2 accepted.)

However, since the atmosphere rotates with the planet, and the apple you mention (at non-zero latitude) is also rotating with the planet and does not suddenly "take off sideways" (i.e. cease its tangential motion) as soon as it starts to fall, it must continue to be subject to centripetal forces, surely?

Isn't this the exact same scenario as a satellite orbiting the Earth, continually falling in a curve towards the Earth. Isn't it 'centrifugal force' that's "keeping it up there"?

Isn't the apple also momentarily "orbiting" - since it has a tangential velocity? Ok, so it's a very rapidly declining orbit, but won't that subtract slightly from g nevertheless? Surely, in order for the apple to fall at exactly g, it must have no tangential motion vector?

[The apple maintains m and v, while r & L are fixed, therefore 'cf' is constant during the fall.]

Which brings me to this:

"

Statement 3 is also correct—it's perfectly possible to imagine a planet that rotates so rapidly that anything not tied down at its equator would be flung off into space. We would say that centrifugal force exceeds the force of gravity, in such a case.note 2"I guess it's okay to imagine it - if your imagination is happy with the irreconcilable physics of it.

A planet of that size spinning that fast would never have formed (to that size) in the first place... due to loss of matter into space. (It rhymes!)

Of course, the real issue with landing on Mars is the lack of usable atmosphere, right? No free braking effect.

«the apple you mention ... is also rotating with the planet and does not suddenly "take off sideways" (i.e. cease its tangential motion) as soon as it starts to fall, it must continue to be subject to centripetal forces, surely?»

I think not, or, if so, it would be its own centripetal force, not that of the planet.

« Isn't this the exact same scenario as a satellite orbiting the Earth, continually falling in a curve towards the Earth. Isn't it 'centrifugal force' that's "keeping it up there"? »

No, centripetal acceleration keeps it in orbit. Since the force actually does work and moves something (the satellite) there's no requirement for an equal and opposite reaction.

« A planet of that size spinning that fast would never have formed (to that size) in the first place... »

Sure, you're right. It's a whadjamacallit, a thought experiment. Thanks for the comments.

...and there's this:

https://mars.nasa.gov/mro/mission/timeline/mtaerobraking/

expat wrote:

"

Since the force actually does work and moves something (the satellite) there's no requirement for an equal and opposite reaction."Huh!? You do mean 'work' in the sense of energy?

If so, that's not correct. Otherwise, you have created a perpetual motion, over-unity machine. Where does the energy to do this work come from? Why does the satellite remain in orbit (unless it's impinging on the atmosphere)?

There is no work, it's simply Conservation of Momentum. (And Conservation of Energy - no energy is being expended.)

And there are, as always, equal and opposing forces. Gravity inwards, and 'centrifugal' force "outwards", in perfect balance. The satellite is not being worked on by either force, it's simply "sliding along" perpendicular to gravity, perpendicular to the 'centrifugal force', on an endless, frictionless elliptical 'track'.

Isn't that correct?

I think you're partly right there. I should not have mentioned work. I just don't see a place for centrifugal force here. Gravity is providing the centripetal force needed to keep Mr. Satellite in orbit, and that's all there is.

I've been thinking about the falling apple—a surprisingly complex problem—and here's what I've concluded:

At the moment the apple detaches from the tree, it has an instantaneous Eastward velocity of 463 m/sec (this tree is at the equator). Per Newton's first law, it will continue that tangential velocity unless acted upon by some diverting force.

It is no longer constrained to follow a curving path, as it was in the Good Old Days when it was attached to the tree. Accordingly, I say that Mr. Apple is not subject to the centrifugal force that the tree and everything else attached to the planet is.ergo, Cotterell is comprehensively wrong.

Hi expat,

Like you, I've also been mulling this problem over further since my previous post.

I agree absolutely, it's a surprisingly complex problem.

And I have to concede, I think you are right. I thought I had you, based on the static variables in the formula quoted, but I made a slight slipup. I took r to be fixed, but that's not really correct. As (the horizontal component of) the apple's velocity is tangential to the surface, r for the apple is increasing. That would result in a decreasing (magnitude-wise, i.e. less -ve) value for 'cf', which doesn't really add up either. So, I think that simply means the formula cannot be applied to an object that is (or would be) moving in a straight line.

So, you are correct! Mr Apple is no longer subject to a centripetal force.

It is a tricky concept. I now also (dis?)agree; the satellite is also not subject to a centripetal force. But, if it was tethered to the planet by a long weightless rope, that was under tension due to holding the satellite in a lower orbit than it would naturally take... But orbiting naturally => no centripetal force...

Ummm... Do you agree?

Another thought!

Is an aeroplane flying through the atmosphere "attached to the planet"?

For example, when performing spiral dives in a small plane. Gut-wrenching... I guess you get used to the feeling, but I never found it at all enjoyable.

I don't think it's the attachment that matters, it's that the object is being forced to turn against "its will" by some external force.

I think the reality is that centripetal force is only imaginary. It's just a variation on change of momentum.

I'm finding it very confusing... Another example, the deflection of electrons by a magnetic field in a CRT. Is this any different from the satellite, or the steel on the edge of a flywheel?

https://en.wikipedia.org/wiki/Centripetal_force

seems to suggest not.

Which brings me back to the naturally orbiting satellite.

No wonder Cotterell doesn't understand it.

« It is a tricky concept. I now also (dis?)agree; the satellite is also not subject to a centripetal force. But, if it was tethered to the planet by a long weightless rope, that was under tension due to holding the satellite in a lower orbit than it would naturally take... But orbiting naturally => no centripetal force...

Ummm... Do you agree? »

So long as "centripetal" is a misprint for "centrifugal", yes.

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