Gravitational Jerk Computation

In summary, In order to lift something at a constant acceleration, you must account for both the force of gravity (G) and the coordinate acceleration (a). Provided that you account for both of these, you can use the equation F=ma to calculate the speed of the object.
  • #1
Will Martin
5
1
I've seen much about jerk, and how it's generally nearly instantaneous, and for general acceleration, that's fine. However, if I lift at a constant acceleration upward slightly stronger than gravity is pulling me downward, the gravitation pull of the Earth will offset part of my force, so that my acceleration upward is lower. Then, as my vessel rises higher, Earth's gravity will be reduced, and will offset less of my force.

For example, if I lift at 1.05G, at the surface, Earth's gravity will reduce my force by 9.80 m/s2, and my resultant force will be 0.05G (0.49 m/s2). However, at an altitude of 160km (low Earth orbit), Earth's gravity will be only 9.33 m/s2, so my resultant force will be 0.96 m/s2. At 400 km altitude (Int'l Space Station), Earth's gravity is reduced further, to only 8.68 m/s2, making my resultant force 1.61 m/s2. If I theoretically lift to geosynchronous orbit, 35,800 km, Earth's pull reduces to 0.22 m/s2. So the difficulty I'm having is a), what is my instantaneous acceleration at specific altitudes (such as LEO or ISS altitudes, or every 1000km up to geo - and I'm pretty sure I have this equation, from sqrt(current acceleration*distance travelled)), and b) how long will it take? I can figure this for consistent acceleration, but as you see, this isn't the case here.

Any help you can give with the equations would be very useful - otherwise, I'll have to resort to fair approximations, which I'd rather not do. And yes, I had calculus, but I haven't used it for several decades ... Thanks ... :-)
 
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  • #2
Will Martin said:
if I lift at a constant acceleration upward

Will Martin said:
what is my instantaneous acceleration

You're contradicting yourself. Please clarify your question and your premise.
 
  • #3
I assure you that if you read the entire premise, the "instantaneous acceleration" at the start isn't static. The force given is 1.05G at lift, but due to the force of gravity of the earth, at sea level, the actual force upward is 0.05G. However, as the craft rises in altitude, the force from Earth reduces, so the resultant force upward grows. No contradiction.
 
  • #4
Will Martin said:
The force given is 1.05G at lift,
1.05G is an acceleration, not a force.

What exactly are you holding constant?
 
  • #5
It sounds as if the distinction between "proper acceleration" and "coordinate acceleration" is crucial here.

Proper acceleration is acceleration that is "felt". You add up all of the external forces that you can feel (e.g. the force of a chair on the seat of your pants) and divide by your mass. The result is your "proper acceleration". Gravity is not a felt force. Instead, it is an inertial force. It does not count toward proper acceleration. [Alternately, you drop a marble so that the marble is in free fall and measure your coordinate acceleration relative to the falling marble. Or you use an accelerometer -- instead of measuring the force on you, you measure the force on a test mass in an accelerometer you are holding]

Coordinate acceleration is acceleration you measure against a particular coordinate system. It is the second derivative of position in the given coordinate system. For instance, in a coordinate system anchored to the Earth.

An object at rest on the Earth's surface has 1 Earth gee of upward proper acceleration and 0 Earth-relative coordinate acceleration.

If one holds proper acceleration constant at 1.05 Earth gees, and starts at rest on the Earth's surface then Earth-relative coordinate acceleration will start at 0.05 gees and increase over time.
 
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  • #6
jbriggs, exactly. I'm trying to determine the Earth-relative speed (thank you for getting my terms straightened out) of a mass as it rises at 1.05G proper acceleration. I've got the formula for my Earth-relative acceleration [1.05G - GravConst*EarthMass*MassObject/(distance from center of Earth)], but the speed is going to be changing and that calculation isn't against the instantaneous acceleration. So for instance, if I am 1000 meters above the Earth's surface (a total of 7378 km between the centers of the objects) and have lifted at a proper acceleration of 1.05G from the surface, what will my speed be? I'm looking for that equation, in relation to distance between the objects …
 
  • #7
anorlunda said:
1.05G is an acceleration, not a force.

What exactly are you holding constant?

Gravity is a force, anorlunda, and 1.05 times gravity is also a force.
 
  • #8
Will Martin said:
Gravity is a force, anorlunda, and 1.05 times gravity is also a force.

They don't have the same units, therefore they're not the same.

However, by Newton's Second Law F=ma, then if we hold mass m constant, then net force F is proportional to acceleration a. In free space, far from any planets or stars, then 0, T is proporitional to A and you can say that a given force implies a given acceleration.

Your scenario sounds like a rocket accelerating directly away from Earth. Let us say that the thrust force of the rocket T, is constant. The gravitational force on the rocket W varies with altitude. W for weight as a function of altitude.

T-W is the net force and T-W=ma.

So, if you hold T constant, the acceleration varies with altitude. If you vary T so that T-W is constant, then acceleration is constant.
 
  • #9
Will Martin said:
Gravity is a force, ...
The strength of the gravitational field is measured in units of acceleration, not force. They are not the same.
 
  • #10
Will Martin said:
Gravity is a force, anorlunda, and 1.05 times gravity is also a force.
Gravity is indeed a force, but a "G" is a unit of acceleration, not a unit of force. So 1.05G is an acceleration.

When dealing with gravity, the mass cancels (the force is ##F=mg##, and when we plug that into ##F=ma## the ##m## cancels from both sides) so the gravitational acceleration is always proportional to the gravitational force. Thus, people often use one when they mean the other and they generally get away with it. You just happened to have chosen a problem in which the difference matters.
 
  • #11
All true, anorlunda, A.T., and Nugatory, and thank you for the discussion of the force of gravity and gravitational acceleration, but still not the answers I need. I am holding T constant, which will cause my resultant acceleration to grow over time. With that in mind, what are my equations for speed and time, given a specific altitude? Am I in the right ballpark with the equation I give above for speed?

[1.05 * Gravitational force - GravConst*EarthMass*MassObject/(distance from center of Earth)]
 
  • #12
Will Martin said:
I am holding T constant, which will cause my resultant acceleration to grow over time. With that in mind, what are my equations for speed and time, given a specific altitude? Am I in the right ballpark with the equation I give above for speed?

[1.05 * Gravitational force - GravConst*EarthMass*MassObject/(distance from center of Earth)]
I see no T in the equation. Also, check if you get the right units from it.
 
  • #13
A.T. said:
I see no T in the equation. Also, check if you get the right units from it.
I introduced T for rocket thrust in #8, trying to get the OP to elaborate.
 
  • #14
Don't make it so complicated. We consider the path of a mass m in Earth's gravity with a constant upwards force of 1.05*g*m where g is the gravitational acceleration at sea level.

The acceleration at a radius r is then ##\frac{a(r)}{g}=1.05-\frac{R^2}{r^2}## where R is the radius of Earth. By definition ##a=\frac{d^2 r}{dt^2}##, so we have a differential equation for the motion.
To find the velocity at a given height conservation of energy can be used. ##\frac{d(v^2)}{dr} = 2a(r)##, integrate to get v2, then take the square root.
To find the time you can integrate over 1/v. It is not guaranteed that this is a nice solvable integral, however.
 
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  • #15
Will Martin said:
All true, anorlunda, A.T., and Nugatory, and thank you for the discussion of the force of gravity and gravitational acceleration, but still not the answers I need. I am holding T constant, which will cause my resultant acceleration to grow over time. With that in mind, what are my equations for speed and time, given a specific altitude? Am I in the right ballpark with the equation I give above for speed?

[1.05 * Gravitational force - GravConst*EarthMass*MassObject/(distance from center of Earth)]
So, Jerk is a rate of change of acceleration. yes, in your set of conditions, acceleration will change with distance from the surface of the earth. (if thrust is kept constant which is near impossible)
 

1. What is gravitational jerk computation?

Gravitational jerk computation is a mathematical calculation that is used to measure the rate of change of gravitational acceleration over time. It is a measure of how quickly the strength of gravity is changing at a specific point in space.

2. How is gravitational jerk computed?

Gravitational jerk is computed by taking the third derivative of the gravitational acceleration equation with respect to time. This involves taking the rate of change of the rate of change of acceleration, resulting in a unit of distance per time cubed.

3. What is the significance of gravitational jerk in physics?

Gravitational jerk is significant in physics because it provides insights into the dynamics of gravitational fields and the behavior of objects in those fields. It can also be used to study the properties of celestial bodies and the nature of space-time.

4. How is gravitational jerk related to other gravitational parameters?

Gravitational jerk is related to other gravitational parameters, such as gravitational acceleration, velocity, and position, through mathematical equations. It is a higher-order derivative of acceleration and is used to calculate changes in these other parameters over time.

5. What are some practical applications of gravitational jerk computation?

Gravitational jerk computation has various practical applications, including spacecraft navigation, orbital mechanics, and gravitational wave detection. It can also be used in the study of planetary motion, gravitational lensing, and the behavior of objects in extreme gravitational fields.

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