# Gravitational field conceptual question

• PrakashPhy
In summary, the conversation discusses a thought experiment where there are only two masses in a universe and how to find their positions as a function of time due to gravitational interaction. The solution involves applying Newton's laws and solving a non-linear ODE, which can be done numerically if not analytically. The conversation also mentions that in real life, the 2-body problem can get more complicated and may require numerical solutions.
PrakashPhy
Suppose a case where two masses are essentially far away form all other bodies so that we can assume the two masses are not affected by any other interactions going on beside them and they do not have the property of gravitation until a time( they are perfectly isolated{in theory everything is possible}). If suddenly at t=0 the gravitational field starts how do we find the position of the two masses as a function of time due to gravitational interaction ( and nothing else)??

PrakashPhy said:
Suppose a case where two masses are essentially far away form all other bodies so that we can assume the two masses are not affected by any other interactions going on beside them
For the sake of the thought experiment, we can imagine a universe with only two masses in it.
and they do not have the property of gravitation until a time( they are perfectly isolated{in theory everything is possible}).
In theory, not everything is possible, that is why we have theories. However, we can imagine that the objects have particular positions at time t=0 when
If suddenly at t=0 the gravitational field starts how do we find the position of the two masses as a function of time due to gravitational interaction ( and nothing else)??

You mean apart from finding their mutual acceleration by Newton's gravitation?

Unless I am missing something in the question, this is a textbook example of a classical mechanics problem. You simply apply Newton's laws, F = m a where the force is given by Newton's law of gravitational. Break vectors into components and integrate the acceleration, which is the second derivative of position with respect to time, to get equations of motion, being careful to remember that the force of gravity depends on position as well. This is all pretty basic stuff you could find in any textbook. Which part don't you understand?

chrisbaird said:
You simply apply Newton's laws, F = m a where the force is given by Newton's law of gravitational. Break vectors into components and integrate the acceleration..

You don't even need to decompose vectors, since there are only two masses in this universe and I assume gravity attracts them "toward" each other.

The resulting differential equation d^2x/dt^2 = K/x^2 (where K is some constant) is not trivial to solve though since it's a non-linear ODE. Unless I'm forgetting something basic.

One can use conservation of energy to get the velocity at any point in the trajectory, but I forget if this greatly simplifies the solution for the full equation of motion.

$$\ddot{x} = \frac{K}{x^2}$$

multiply both sides by $\dot{x}$
$$\dot{x}\ddot{x} = \frac{K\dot{x}}{x^2}$$

Now you can integrate both sides, rearrange to get:

$$\dot{x} = \sqrt{ \frac{2K}{x} +C }$$

which you can do. It's not pretty.

$$t = \frac{\dot{x}}{C} - \frac{K\ln|2K + 2C\dot{x}|}{C^{3/2}} + D$$

More generally:
http://www.berkeleyscience.com/pm.htm

Matterwave said:
The resulting differential equation d^2x/dt^2 = K/x^2 (where K is some constant) is not trivial to solve though since it's a non-linear ODE. Unless I'm forgetting something basic.

I meant setting up the problem is straight-forward. Solving the differential equation is not always doable analytically. When not, you an always solve differential equations numerically, either by using per-packaged software or writing your own code.

Oh boy - IRL: differential equations can get very nasty very fast. Real life is messy.

The nice quadratic ballistic equations we learn in school only work under special conditions ... linear gravity and no other forces.

Actual accurate satellite orbits are determined numerically and cannot be expressed analytically like this.
OTOH: the 2-body problem described was worked out quite a while ago.

Can't the two body problem be reduced to a similar one body problem? Taking the Lagrangian of the system, assuming the origin to be the COM ($r_1m_1 = -r_2m_2$) and defining $r = r_1 - r_2$you have
$$L = \frac{1}{2}m_1\dot{r_1}^2 + \frac{1}{2}m_2\dot{r_2}^2 - U(r)$$
We can express the position vectors as $r_1 = \frac{m_2r}{m_1+m_2}$ and $r_2 = \frac{m_1r}{m_1+m_2}$, then define $m = \frac{m_1m_2}{m_1+m_2}$ which then reduces the Lagrangian to
$$L = \frac{1}{2}m\dot{r}^2 - U(r)$$
Where
$$U(r) = -\frac{m_1m_2 G}{r}$$

I guess this would be more of a question than an answer though? :)

Last edited:

## What is a gravitational field?

A gravitational field is a region in space where an object with mass experiences a force due to the gravitational pull of another object. This force is the result of the curvature of space caused by the presence of mass.

## What is the difference between gravitational field and gravitational force?

The gravitational field is a measure of the force per unit mass at a certain point in space, while the gravitational force is the actual force acting on an object with mass. The gravitational field is a vector quantity, while the gravitational force is a scalar quantity.

## How does the strength of a gravitational field change with distance?

The strength of a gravitational field decreases with distance. This is because the force of gravity is an inverse-square law, meaning that as distance increases, the force decreases by the square of the distance.

## Can gravitational fields cancel each other out?

Yes, gravitational fields can cancel each other out. This is known as destructive interference, where two fields with equal magnitude but opposite direction cancel each other out resulting in a net force of zero.

## How does mass affect the strength of a gravitational field?

The strength of a gravitational field is directly proportional to the mass of the object creating the field. This means that the larger the mass, the stronger the gravitational field it produces.

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