# Calculating the time it takes for something to fall over long distances

1. Oct 26, 2008

### hover

1. The problem statement, all variables and given/known data

This isn't really a homework question that I have but more of a question i am trying to find out myself. When objects are near the surface of the earth, objects accelerate at an average of 9.8 meters per second^2. However when we get farther away from the earth, say have the distance it takes to get to the moon, the acceleration is different. Not only that but as the object gets closer to the earth the acceleration increases and can not be treated as a constant anymore. I've been thinking in order to resolve this I must try to make an equation with change in acceleration, a jerk. I want to see what you guys think there are any flaws in my process. I also want to figure out the amount of time from my final equation as the title implies.

2. Relevant equations
$$A= \frac{\ a+ao}{2}$$

$$j= \frac{\ a-ao}{t}$$

$$d= vot+.5At^2$$

Where A is the average acceleration, a is the final acceleration, ao is the initial acceleration, j is the average jerk, t is time, d is the amount of displacement and vo is the initial velocity.

3. The attempt at a solution

$$d= vot+.5At^2$$

I notice that it has the variable A which needs to be taken out. So I replace A with this equation-

$$A= \frac{\ a+ao}{2}$$

and it becomes

$$d= vot+.5(\frac{\ a+ao}{2})t^2$$

Next i need to decide to get rid of a or ao so i can somehow put j in there. It doesn't really matter but i decide to replace a with this equation

$$j= \frac{\ a-ao}{t}$$
$$jt= a-ao$$
$$jt+ao=a$$

So once again the new equation is this-

$$d= vo+.5(\frac{\ jt+ao+ao}{2})t^2$$

Now i go through the equation and simplify

$$d= vot+.5(\frac{\ jt+2ao}{2})t^2$$
$$d= vot+.5(\frac{\ jt^3+2aot^2}{2})$$
$$d= vot+(\frac{\ jt^3+2aot^2}{4})$$
$$d= \frac{4vot}{4}+\frac{\ jt^3+2aot^2}{4}$$
$$d= \frac{\ jt^3+2aot^2+4vot}{4}$$

this is what i get for a final equation
$$d= \frac{\ jt^3+2aot^2+4vot}{4}$$

This equation makes sense because when j is 0 you get the original equation of
$$d= vot+.5At^2$$
and ao becomes A because there is no change in acceleration.

However I don't know if my final equation is correct for sure and I want some feedback to see if it is right. If it is right then we can move on to the question of finding out what t is when all other variables are known. The problem is I don't know how to solve for t. My mind says to make one side of the equation 0 so

$$0= \frac{\ jt^3+2aot^2+4vot-4d}{4}$$

And i guess i can get rid of the 4 in the denominator by multiplying both sides of the equation by 4, giving me this

$$0= jt^3+2aot^2+4vot-4d$$

But then what? I can't use the quadratic formula. I could factor like this

$$0= t(jt^2+2aot+4vo)-4d$$

but that doesn't really get me to far. Can someone help, please?
Thanks for helping me out.

2. Oct 26, 2008

### hover

Does anyone want to help me? Your help is appreciated.

3. Oct 26, 2008

### hover

Can someone please help? I want to try to figure out what t is in my final equation (if it is correct) and i can't do it. I don't know how to do it.

4. Jan 31, 2009

### Edward G

I think you would need calculus to do this properly, specifically that

$$Force = \frac{d(mv)}{dt}$$

and the force if the force is just gravitational from a single body,

$$Force = \frac{Gm_1 m_2}{r^2}$$

G - Newton's Gravitational constant
m1 and m2 are the masses of the bodies being attracted.

This comes purely from the fact that the acceleration isn't constant, but it is changing all the time.

5. Jan 31, 2009

If you want to calculate the speed at any point you could use potential energy lost = kinetic energy gained.For large distances where we have to take into account the reduction of g the P.E. equation is given by:
P.E.=GMm(1/a minus1/b)
G= universal gravitational constant
M= mass of earth
m= mass of falling object
a= radius of earth(or the distance from the earths centre that the object falls to)
b= original distance.
For a problem where b is very large compared to a we can usually assume that 1/b is negligible compared to 1/a.If my memory serves me correctly the smaximum speed reached will be about 11 km/s

6. Jan 31, 2009

### Nabeshin

P.S: You yourself noted that acceleration is constant. You cannot use the basic equations of kinematics with non-constant acceleration, so the first step of your derivation is incorrect. As was noted by Edward G, you need calculus to arrive at and solve a differential equation to get the answer.