Free fall trajectory in non constant gravitational field

In summary, you need to integrate the distance/acceleration formula taking into account the change in g to get the total distance traveled. The method and final formula is found by substituting Newton's law of gravitation in for the constant g.
  • #1
Creator
566
6
Rusty on the math...
I am working on a problem and need the derivation of a free fall trajectory for an object at a distance above Earth where the change in acceleration is not negligible. How do I integrate the distance/ acceleration formula taking into account the change in g to get the total distance traveled.
The method and final formula...?
I'm looking for total distance after time t, trying to see how it scales as what power of t.
thanks.
 
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  • #2
You need to know what the function of g with respect to position is, and then plug it into your integral instead of assuming it is a constant.
 
  • #3
Let me get this right. The object is tracing a trajectory where 'g' changes according to the distance from the Earth?
 
  • #4
siddharth23 said:
Let me get this right. The object is tracing a trajectory where 'g' changes according to the distance from the Earth?

Over long enough range, g is not constant as the Earth is not of uniform density. Although I suspect what the OP was interested in was the decreasing effect of gravity with altitude. In which case you would substitute Newton's law of gravitation in for the constant g in your integral g(x)=G M1 M2 / x^2.
 
  • #5
You possibly can't determine the density at every point. Has to be g(x). Like QuatumPion said, put it in the equation in terms of x and integrate, In this case, it'll be in terms of 'y' as 'x' would be the horizontal distance traveled by the projectile.
 
  • #6
siddharth23 said:
You possibly can't determine the density at every point. Has to be g(x). Like QuatumPion said, put it in the equation in terms of x and integrate, In this case, it'll be in terms of 'y' as 'x' would be the horizontal distance traveled by the projectile.

Probably, but even for the first case if you have a gravity map and know the trajectory/inclination than you could make a function of g with respect to position, although that would be fairly complicated (depends on orbital parameters, rotation of earth, etc).
 
  • #7
Creator said:
Rusty on the math...
I am working on a problem and need the derivation of a free fall trajectory for an object at a distance above Earth where the change in acceleration is not negligible. How do I integrate the distance/ acceleration formula taking into account the change in g to get the total distance traveled.
The method and final formula...?
I'm looking for total distance after time t, trying to see how it scales as what power of t.
thanks.
Sidhartha23 and QuantumPion have argued back and forth but we haven't heard back from Creator so I am going to look at the specific case Creator asked about: An object falling from such a great height the we must use the general law for gravity: [itex]F= -GmM/r^2[/itex], rather than treating F as a constant, still satisfies "Force= mass times acceleration". Here "m" is the mass of the object, M is the mass of the Earth so we have [itex]ma= -GmM/r^2[/itex] or
[tex]a= \frac{dv}{dt}= -\frac{GM}{r^2}[/tex]

We would like to integrate that but we need to integrate with respect to t and have "r" on the right. To fix that, we can use a method called "quadrature". By the chain rule, dv/dt= (dv/dr)(dr/dt)= v(dv/dr) becaues v, the speed, is dr/dt. Now we have
[tex]v\frac{dv}{dr}= -\frac{GM}{r^2}[/tex]
which we can write, in "differential form" as
[tex]vdv= -\frac{GM}{r^2}dr= -GMr^{-2}dr[/tex]

Integrating both sides,
[tex]\frac{1}{2}v^2= GMr^{-1}+ C[/tex]
assuming that the object falls from a standstill at height R, v= 0, so [itex]0= GMR^{-1}+ C[/itex], [tex]C= -GMR^{-1}[/tex].

From that,
[tex]v= \frac{dr}{dt}= \sqrt{2GMr^{-1}- 2GMR^{-1}}= \sqrt{2GM}\sqrt{r^{-1}- R^{-1}}[/tex]

We can write that as
[tex]\frac{dr}{\sqrt{2GM}\sqrt{r^{-1}+ R^{-1}}}= dt[/tex]

Integrating both sides of that will give t as a function of r which can then solved for r as a function of t. However, the left side cannot be integrated in terms of "elementary functions". It is an "elliptic integral" (the name, unsurprisingly, connected with the fact that planets move in ellipses around the sun). There used to be multi-volumes of tables of elliptic integrals.
 
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  • #8
Thanks for all the quick responses...
You are correct Siddarth23 and Quantum Pion...about my question.
Creator
 
  • #9
HallsofIvy said:
Sidhartha23 and QuantumPion have argued back and forth but we haven't heard back from Creator so I am going to look at the specific case Creator asked about: An object falling from such a great height the we must use the general law for gravity: [itex]F= -GmM/r^2[/itex], rather than treating F as a constant, still satisfies "Force= mass times acceleration". Here "m" is the mass of the object, M is the mass of the Earth so we have [itex]ma= -GmM/r^2[/itex] or
[tex]a= \frac{dv}{dt}= -\frac{GM}{r^2}[/tex]

We would like to integrate that but we need to integrate with respect to t and have "r" on the right. To fix that, we can use a method called "quadrature". By the chain rule, dv/dt= (dv/dr)(dr/dt)= v(dv/dr) becaues v, the speed, is dr/dt. Now we have
[tex]v\frac{dv}{dr}= -\frac{GM}{r^2}[/tex]
which we can write, in "differential form" as
[tex]vdv= -\frac{GM}{r^2}dr= -GMr^{-2}dr[/tex]

Integrating both sides,
[tex]\frac{1}{2}v^2= GMr^{-1}+ C[/tex]
assuming that the object falls from a standstill at height R, v= 0, so [itex]0= GMR^{-1}+ C[/itex], [tex]C= -GMR^{-1}[/tex].

From that,
[tex]v= \frac{dr}{dt}= \sqrt{2GMr^{-1}- 2GMR^{-1}}= \sqrt{2GM}\sqrt{r^{-1}- R^{-1}[/tex]

We can write that as
[tex]\frac{dr}{\sqrt}2GM}\sqrt{r^{-1}+ R^{-1}= dt[/tex]

Integrating both sides of that will give t as a function of r which can then solved for r as a function of t. However, the left side cannot be integrated in terms of "elementary functions". It is an "elliptic integral" (the name, unsurprisingly, connected with the fact that planets move in ellipses around the sun). There used to be multi-volumes of tables of elliptic integrals.
Thank you Halls ...that is exactly what I was looking for...however, I only followed you up to this eqn. (4th one)...
\frac{1}{2}v^2= GMr^{-1}+ C

And your next step (5) I don't understand since I thought the v here was final velocity...

Your last two eqns. remained in ITEX form and did not print correctly...could you give the last two again. Maybe its my browser...mot too famailiar with mathjax.
Thanks;
Creator
 
Last edited:

What is free fall trajectory in a non constant gravitational field?

Free fall trajectory in a non constant gravitational field refers to the path that an object takes when it is falling under the influence of a gravitational force that is not constant. This can occur when the strength of the gravitational force changes due to varying distances from the center of mass or when there are multiple gravitational forces acting on the object.

What factors influence the trajectory of an object in free fall in a non constant gravitational field?

The trajectory of an object in free fall in a non constant gravitational field is influenced by the strength and direction of the gravitational force, the mass and initial velocity of the object, and the shape and composition of the object.

How can the trajectory of an object in free fall in a non constant gravitational field be calculated?

The trajectory of an object in free fall in a non constant gravitational field can be calculated using mathematical equations such as Newton's laws of motion and the law of universal gravitation. These equations take into account the various factors that influence the trajectory of the object and can be solved using numerical methods or computer simulations.

What are some real-world examples of free fall trajectory in a non constant gravitational field?

Some examples of free fall trajectory in a non constant gravitational field include a spacecraft orbiting a planet, a satellite orbiting Earth, and a comet falling towards the Sun. In all of these cases, the gravitational force acting on the object is not constant, resulting in a non-linear trajectory.

How does free fall trajectory in a non constant gravitational field differ from free fall in a constant gravitational field?

In free fall in a constant gravitational field, the gravitational force acting on the object remains constant throughout the fall, resulting in a parabolic trajectory. However, in a non constant gravitational field, the gravitational force changes, causing the trajectory to deviate from a parabolic path and follow a more complex path. Additionally, the speed and acceleration of the object will also vary in a non constant gravitational field.

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