Solving Differential Equations: Falling Objects & Linear Gravity

[DonDiablo]

Hy folks,

Upfront I want to apologize for my writing and my dissability to use correct symbols to ease readability of the example. Ok now that that's done I just want to start upfront.

If we set a usual example of an object falling from a tower with a height of x meters and assume that the gravitational constant is the same during the way of falling - then the velocity is given by: v=a*t and integrated over time the way is given by s = (a*t²)/2 - so if you wanted to know how long a given object would fall down until it reached the ground you'd transform the equation and solve it.

But in a hypothetical scenario where "g" was not constant but would linearly grow the nearer the object came to the ground during the process of falling equations would look different. So that is where my problem starts. I am for some reason (the obvious reason of course being me being stupid) not able to create the right equations to be able to adequately express the premise.

(Obv. I'm a relative novice in the field of differential equations)

Edit: in this hypothetical world "g" would obv. be the strongest basically on the surface of the earth)

Thx upfront for the help!
kind regards
Don

 

[sophiecentaur]

So that is where my problem starts. I am for some reason (the obvious reason ofc being me being stupid) not able to create the right equations to be able to adequately express the premise.

In situations where g can't be considered to be constant, the Maths gets more complicated and Calculus is needed. That could be your problem. From the surface of the Earth to a considerable height, g can be considered to be practically the same. When you come to dealing with orbits - like the Moon, the acceleration takes the form
a = MG/d²
where M is the mass of the Earth (or any other large mass) and G is the Gravitational Constant (not g!)
The actual force F on a mass m is ma, so
F = mMG/d²

The 2D map at the top of this link will let you browse around the topic without trudging through a textbook.
If you want to work out the initial and final speeds of an object, falling directly to a planet, you just need to integrate the acceleration over the distance fallen. This relates to the change in Potential Energy - see the link.

 

[jbriggs444]

(Obv. I'm a relative novice in the field of differential equations)

In the general case, one might have to deal with differential equations. This situation is simple enough to avoid that.

In this scenario, the force on a falling body is a function of position alone. That means that you can take the integral of force times distance and arrive at a figure for energy gained falling along a path. Integrating the dot product of instantaneous vector force times incremental vector displacement along a path is called taking a "path integral".

With the above result in hand, you would be in a position to know the kinetic energy of the falling object at every point along its path. That means that you could easily solve for its speed at every point. [$E=\frac{1}{2}mv^2$, solve for $v$].

If you wanted to calculate the total time to fall, you could integrate a second time. This time you would be after the integral of time taken per unit distance [$\frac{1}{v}$] times incremental distance along the path.

 

[DonDiablo]

Hi and thank you guys for your quick responses! I will respond to each one of you in the order you replied to my original post!

@sophiecentaur: Yes- I understand your point and I do understand that in a case where I want to create a scenario in which "g" would not be constant given the position of the falling body I would have to define a law or at least a field of force that gives the rules of that scenario the simplest probably being the one where you apply the gravitational law (is it even called like that?)... where F = mMG/d²... thus -given how big M (the mass of the of the earth) is the the change of F (and thus a - the "acceleration component of F (on the body whose mass is to be considered constant during the process of falling) is relatively small it is usually considered to not be worth to be considered (wow - great formulation i know *rolleyes)...

@jbriggs444 I am not sure if I am able to follow every aspect of your post, but it makes sense to me that the energy of an object is a function of its position. I also understand that the gravitational field can be considered a vector field in this case (well if we'd be super precise not really) with all vectors at any given position pointing in the same directions. So I assume that's what you mean with instantaneous vector force... hmm ok... but i get that the result of the equation would turn out to be rather simple since the path we would be integrating over is just a straight line in space... (of course we are not "going Einstein" here)...

Additional note: I honestly blame my imagination for causing most of the problems I am having with fairly simple set ups... and also I mess up so man things in my head! But the impulses given by your responses have been very helpful!
Kind regards
Don

 

[jbriggs444]

but i get that the result of the equation would turn out to be rather simple since the path we would be integrating over is just a straight line in space...

Right. In general it could be a vector field and a complicated path. But in the case at hand, the path is straight down, the force is straight down and you can just evaluate the integral: $\int_{top}^{bottom} f(y) \ dy$