Check out some similar questions!

survivorboi · Jul 27, 2009, 02:25 PM

How do I use Newton’s F= G{(M*m) / r^2} to figure out an objects freefall?

An object is throwing straight up so that it returns exactly 5 days later. What would be the velocity needed? (Like how fast does the object have to be thrown up)

I want to use the equation above to solve it!

ebaines · Jul 28, 2009, 01:41 PM

Sorry, but you're not going to find a nice close-formed solution for this! The equation of motion is:

$ a(t) = \frac {dv} {dt} = - \frac {GM} {x^2} $

where x is function of t, and hence a is not constant.

$ V = \int a(t) dt \ = \ \int - \frac {GM} {x^2} dt \ = \ \frac {dx} {dt}\\ x = \int v(t) dt \ = \int ( \int - \frac {GM} {x^2} dt ) dt $

At this point I get stuck!

For the case where Gm/x^2 is essentially a constant = g (near the earth's surface), then a as a constant and this becomes simply:

$ x = \int v(t) dt \ = \int ( \int ( - g )dt ) dt = \int (-gt +V_o)dt = - \frac 1 2 g t^2 + V_0 t + x_0. $

This is the quadratic form for motion under constant acceleration that is so familiar to physics students. The round trip time for the object to go from $x_0$ = 0 back to 0 is:

$ 0 = - \frac 1 2 g t^2 + V_0 t + 0 \\ t = \frac {2 V_0} g $

survivorboi · Jul 28, 2009, 05:32 PM

$ a(t) = \frac {dv} {dt} = - \frac {GM} {x^2} $

So that equation above is the actual equation, right? Because you wrote a bunch of numbers and I don't know which is which.

Can you please tell me how to substitute the letters for numbers according to the problem I listed above? Thanks

Please make it as simple as possible because I'm only in Algebra (8th grade)

ebaines · Jul 29, 2009, 05:46 AM

Originally Posted by survivorboi

$ a(t) = \frac {dv} {dt} = - \frac {GM} {x^2} $

So that equation above is the actual equation, right? Because you wrote a bunch of numbers and I don't know which is which.

Can you please tell me how to substitute the letters for numbers according to the problem I listed above? thanks

Please make it as simple as possible because I'm only in Algebra (8th grade)

Yes, that is the basic equation, which governs the object's acceleration, a. The acceleration over time governs the object's velocity v, and its velocity over time determnes its position x. But note the acceleration is dependent upon the object's distance from the center of the earth, x - hence you get a set of equations that is circular and very difficult to solve - you have to know the object's position to determine its acceleration, and its acceleration to determine its position. So there is no simple formula that you can plug terms into that would let you calculate the time it takes for an object to come back to earth after being launched, except for the case where the object is in a uniform gravitational field. That equation is

$ t = \frac {2 V_0} g $

where $V_0$ is the object's initial velocity upward and g is the acceleration due to gravity at the earth's surface: g = 9.8 m /sec^2, or 32.2 ft/sec^2. This equation is suitable for things like baseballs, rocks, etc - anything that won't go any higher than a few hundred miles and doesn't go into orbit. As an 8th grade algebra student, I suggest that you work with this equation first, and not worry about the calculus concepts.

survivorboi · Jul 29, 2009, 09:29 AM

Originally Posted by ebaines

Yes, that is the basic equation, which governs the object's acceleration, a. The acceleration over time governs the object's velocity v, and its velocity over time determnes its position x. But note the acceleration is dependent upon the object's distance from the center of the earth, x - hence you get a set of equations that is circular and very difficult to solve - you have to know the object's position to determine its acceleration, and its acceleration to determine its position. So there is no simple formula that you can plug terms into that would let you calculate the time it takes for an object to come back to earth after being launched, except for the case where the object is in a uniform gravitational field. That equation is

$ t = \frac {2 V_0} g $

where $V_0$ is the object's initial velocity upward and g is the acceleration due to gravity at the earth's surface: g = 9.8 m /sec^2, or 32.2 ft/sec^2. This equation is suitable for things lik baseballs, rocks, etc - anything that won't go any higher than a few hundred miles and doesn't go into orbit. As an 8th grade algebra student, I suggest that you work with this equation first, and not worry about the calculus concepts.

Velocity, is that talking about any type of speed? Like a baseball going at 70 miles per hour, is it's velocity, right?

You tossed the baseball straight up at 70 miles per hour, going for 5 seconds.

So then, V^o would be 70 milers per hour (you probably have to convert it to meters to comply with g)

So 2 V^o = 2(70)

Divide by 9.8m/sec^2

How is that possible? How can you divide seconds by meter? It's not like terms. You can't add 3x and 3y. How can you divide 9.8m and seconds?

ebaines · Jul 29, 2009, 09:51 AM

Velocity is indeed speed, except when we use the word velocity instead of speed it means that we're talking about the object's direction of motion as well as how fast it goes. Consider a ball tossed in the air - to figure out how long it taes to fal back to earth it's important to know the speed that it's thrown as well as its direction - if you throw it straight up at 70 MPH it will take longer to reach the ground than if you throw it straight down. So the $V_0$ term in the formula is strictly speaking the magnitude of the object's initial velocity in the upward direction.

Speed (or velocity) has units of distance per unit time - like MPH, or feet per second, which we write as ft/sec. Acceleration is a measure of how fast velocity is changing, and so has units of velocity change per second, or ft per second per second, or feet per second squared, written as ft/sec^2.

The units do indeed work out - if you divide velocity (feet per second) by acceleration ( ft per second squared) you are left with seconds. Using simple algebra:

$ \frac { \frac {ft} {sec} } { \frac {ft} {sec^2}} = \frac {ft} {sec} \cdot \frac {sec^2} {ft} = \frac {\cancel {ft}} {\cancel {sec}} \cdot \frac {sec^{\cancel{2}}} {\cancel{ft}} = sec $

So to use the formula: suppose you toss a ball upward at 70 MPH. First convert that to either ft/sec of m/s; I like feet/sec:

70 Miles/Hr * 5280 ft/Mile * 1 Hr/3600 Sec = 102.67 ft/sec.

The time it takes for the ball to fall back to earth is:

$ t = \frac {2* 102.67 ft/sec } {32.2 ft/sec^2} = 6.38 sec $

See how the units work out?