# I Gravitational Jerk Computation

#### Will Martin

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 ... :-)

Related General Physics News on Phys.org

#### anorlunda

Mentor
if I lift at a constant acceleration upward
what is my instantaneous acceleration

#### Will Martin

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.

#### anorlunda

Mentor
The force given is 1.05G at lift,
1.05G is an acceleration, not a force.

What exactly are you holding constant?

#### jbriggs444

Homework Helper
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.

Last edited:

#### Will Martin

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 …

#### Will Martin

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.

#### anorlunda

Mentor
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.

#### A.T.

Gravity is a force, ...
The strength of the gravitational field is measured in units of acceleration, not force. They are not the same.

#### Nugatory

Mentor
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.

#### Will Martin

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)]

#### A.T.

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.

#### anorlunda

Mentor
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.

#### mfb

Mentor
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.

#### zanick

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)

"Gravitational Jerk Computation"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving