[I]True[/I] Gravitational Acceleration

[zking786]

While we are told that g = 9.81 m/s/s, this is only an approximation and isn't necessarily always true when we do precise calculations. Since, by the Universal Gravitation Laws, acceleration is determined by the mass of each object and the distance between the objects, we can't necessarily use a "fixed" gravitational acceleration of 9.81, since as we are free-falling gravity's acceleration changes as a function of distance. In fact, even mass changes as we accelerate. I've always wondered how we could compute the true free-fall for an object of mass m as it falls.

 

Any ideas?

[the tree]

With great difficulty, you'd need to work out a function for how the gravitational attraction (and thus accelleration) increases as the object falls. Then you'd want to factor in relativity and the such like. Then it should be about accurate but unlikely that different from a constant [math]9.81ms^{-2}[/math] anyway.

[Klaynos]

You'd use: [math] g=- \frac{GM}{r^2}[/math]

[the tree]

...or that of course. Not actually studying physics, my only understanding comes from mechanics classes.

 

Klaynos, I don't suppose you could tell us what that means?

[Sisyphus]

G is the gravitational constant, M is the mass of the attracting body, and r is the distance between the two centers of mass. It's negative because you're accelerating "downwards," or towards each other.

[zking786]

The universal gravitation law makes sense, but how do we use calculus to say... graph the equation or find the position data?

 

How does this take relativity into account? Any idea on how we could use differential equations to compute?

[timo]

I don´t have much time right now, so I´ll stick to the questions that can be answered quickly:

How does this take relativity into account?

Above? Not at all.

Any idea on how we could use differential equations to compute?

There´s a pretty wide range of methods to solve differential equations numerically. The easiest one is the following:

1) Start with a starting position x(0) and a starting velocity v(0).

2) chose a sufficiently small timestep dt.

3) Set t = 0.

4) x(t+dt) = x(t) + v(t)*dt, v(t+dt) = v(t) + g(x)*dt where g(x) might be the gravitational acelleration posted by Klaynos, for example (depeds on the system)

5) t = t+dt

6) goto 4 until you reached a t which is sufficiently large for your purposes.

 

Or did you mean how the movement equation looks like as a differential equation? That´s d²x/dt² = g(x).

 

Note that Klaynos and the follow-up poster(s) assumed that g is negative. That´s true if one assumes only a 1D problem with the positive direction beiong defined as "away from the attracting mass". But in general, the acelleration g is a vector anyways and for simple problems you know where it points - therefore knowing the magnitude (the positive value) is sufficient.

[zking786]

Simply stated, I'm wondering how could I derive a position vs. time graph? What would the equation be?

 

How would I take both relativity and the gravitational laws into account and compute the two differential equations?

[zking786]

B U M P -- Anyone still here?

[the tree]

Remembering that I'm not that good with mechanics, I think it'd be something like this.

[math]t=\int_{r}- \frac{GM}{r^2}[/math]

That's not involving relativity, but considering that relativity is only mesurable with huge things, I don't think it'd matter.

It also ignores terminal velocity, for simplicities sake.

[Sisyphus]

Right, it's just unhindered acceleration, so other factors like wind resistance are ingnored. For relativity I think you'd have to specify a reference frame.

[zking786]

the tree:

How did you derive that integral? Also, this might sound extremely elementary, but why don't you have an upper bound on the integral? Also, since r varies with time, how can you have it as one of the bounds. Important thing to consider is I only have one year of calculus (brief differential eqns unit). Would appreciate clarification.