Can Non-Constant Gravity Affect Velocity and Acceleration Equations?

In summary, the speaker is trying to create an equation that describes distance, velocity, and acceleration for a free falling object with non-constant acceleration due to changing gravity. They are also looking for a way to describe the object relativistically without its velocity exceeding the speed of light. They mention a related issue of falling into a black hole and provide links to previous discussions on the topic.
  • #1
gill12
2
0
I am trying to devise an equation to describe distance, velocity, and acceleration with non-constant acceleration due to gravity changing with respect to distance. For instance, at 100,000 meters the gravitational acceleration is let's say X, yet at 1,000 meters the gravitational acceleration is 10,000X. This gives different distance, velocity, and acceleration equations. Can anyone help me? Thanks in advance!
 
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  • #2
I am trying to calculate the distance, velocity, acceleration, and time of a free falling object. Yet, my velocity after a certain amount of time and/or sufficient acceleration will exceed c, 299792458 m/s. I am looking for a transform for any of these 4 parameters that will describe the object relativistically. Basically, what transform do I use and what do I use it on so that my velocity, v, does not exceed c, but only approaches it as time goes to infinity?
 
  • #3
You might want to look at http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

Note that this is a rocket with a constant proper acceleration of 1g. There isn't any way for a rocket to have a constant coordinate acceleration of 1g in flat space-time.

A somewhat related question is the issue of falling into a black hole. This may or may not be closer to what you are interested in, but it is complicated by the presence of curved space-time and its associated metric coefficients.

This gets involved, but has been discussed (if that's what you're interested in), see for instance the following posts (and the associated threads).

https://www.physicsforums.com/showpost.php?p=1209950&postcount=70
https://www.physicsforums.com/showpost.php?p=602558&postcount=29
 
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1. What is a transform for velocity?

A transform for velocity is a mathematical operation that is used to convert between different units of velocity. It allows us to express velocity in terms of different measurement units, such as meters per second, kilometers per hour, or miles per hour.

2. Why do we need a transform for velocity?

We need a transform for velocity because different units of measurement are commonly used to express velocity, and it is often necessary to convert between these units for various calculations or comparisons. A transform for velocity makes it easier to work with different units of velocity by providing a standard method for conversion.

3. How is a transform for velocity different from a conversion factor?

A transform for velocity and a conversion factor serve similar purposes, but they are not the same. A conversion factor is a numerical value that is multiplied by a quantity to convert it from one unit to another. A transform for velocity is a mathematical equation that takes into account not only the numerical value, but also the units and their relationships to each other.

4. Are there different types of transforms for velocity?

Yes, there are different types of transforms for velocity depending on the units being converted. Some common transforms include converting between meters per second and kilometers per hour, or between miles per hour and feet per second. The specific type of transform used will depend on the units being converted and the desired accuracy.

5. How is a transform for velocity calculated?

A transform for velocity is typically calculated using a formula that takes into account the numerical value and units of the velocity being converted. For example, to convert from meters per second to kilometers per hour, the formula is: km/h = (m/s) x 3.6. This means that to convert a velocity from meters per second to kilometers per hour, you would multiply the value in meters per second by 3.6.

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