Is the gravity model taught in physics and calculus truly quadratic?

Question

First, my apologies to Steve C; I thought that I awarded you the Best Answer for the question on complex numbers in aerospace engineering, but an email said otherwise.  Yours was truly great; I'm new at this, but I'll work on it.

Now, the question.  I've taught precalculus and shown the relation between acceleration, velocity and displacement.  I know that gravity is constant, so it makes sense that acceleration due to gravity is constant, the change in velocity is linear, and the displacement is a quadratic function...

Until...one considers the impact of Newton's Law of Gravitation.  This is what throws me.  We all know that F = ma, but that F is Gm1m2/d^2 in planetary terms.  It's that d^2 that gets me.  Using Earth as being 6373 km in radius, if I dropped Baumgartner-style to Earth from that altitude, wouldn't the acceleration be 1/4 what it is on the surface?  Shouldn't it be about -2.45 m/sec^2?  If that's the case, then don't we have to rethink our model?  It's no longer quadratic, but what is it?  Is the acceleration model really (gulp) a differential equation, since acceleration would depend on distance from Earth, but distance depends on which way we're moving, and that on how we're accelerating?

See the quandary I'm in?  I may be missing something, but I know someone can clarify this.  Who's got it in them to help me out?

Dr. Zorro · Accepted Answer

Indeed, when not approximating GM/r^2 by GM/R^2  (assuming the distance from the center of the Earth say is approximately R, i.e. setting g(r) = g(R) = constant) , we do not have the luxury of a simplified equation of motion  with constant acceleration (m d^2 r /dt^2 = - m g).

It is the constant acceleration that upon integration gives v(t) = v(0) + a t , whilst one further integration gives s(t) = s(0) + v(0) t + 1/2  at^2 .

So, what type of free-fall results when taking the variation of the force into account?
Well, Newton's second law now gives for the acceleration d^2r/dt^2 the differential equation:

m d^2 r(t) /dt = - GMm/r(t)^2

Solving this will give the position as a function of time. Clearly we will not obtain the simple t^2 law. Instead you will find that the position r = r(t) satisfies

(  r0 Sqrt[r0 - r] arctan[Sqrt[r]/Sqrt[r0 -r]]  - (r0 - r)  )/Sqrt[r0 - r] = Sqrt[GM/r0] t

Steve4Physics · Answer

"I know that gravity is constant"
No it isn't - that's just an approximation near the earth's surface, which is adequate for many problems and is use in teaching at an introductory level.

You are correct: displacement being quadratic with time is only an approximation. It is only true in a uniform gravitational field.   As if we deal in large height changes we can no longer use this.

There are different mathematical methods needed.  For simple problems we can use the concept of gravitational potential (V = -GM/d). For more complex problems we need differential equations - but some are not that hard when you have been taught, so no need to gulp!.

In general you need vector calculus - which CAN get a bit messy. This lets you work out the details of elliptical motion of planets around a star for example.

(If you want to be really, really accurate, you can't even use the inverse square law (F=Gm1m2/d^2 ). You have to use General Relativity which is very advanced and way more complicated mathematically.)

or

OldPilot · Answer

No.  Newton is idealized as 2 point masses alone in the universe. With 2 masses acting on a 3rd it becomes the Inverse Cube Law.   After that with more masses it becomes General Relativity and requires 4D Tensors

However, for most problems Newton is close enough and sufficient.  NASA uses Newton to plot spacecraft motion. The rest is not worth the extra computer time

Is the gravity model taught in physics and calculus truly quadratic?

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