# Simple Gravity Question

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alexchamp29
Warning: I'm a freshman chemistry student. My math skills are elementary at best.

Image a ball in deep space. A baseball sounds nice. Assume this ball had no initial velocity but suddenly finds itself caught in the sphere of influence of a massive body. A black hole sounds fun. The ball begins accelerating towards the black hole. What equation, given the mass of ball, the black hole, and initial distance, could yield the ball's velocity at a given distance in its descent?

Homework Helper
Hello Alex, At school you must have done the SUVAT equations where the attractive gravitation acceleration ##g## is a constant, right ?

More in general you have Newton equations governing the motion.

alexchamp29
I haven't taken any physics outside of high school. I do know that the acceleration of the object is it's force over its mass. And the force acting on it is "Gmm/r^2". I guess I just need to know how to find its acceleration as "r" gets lower? But how fast "r" changes depends on the acceleration of the object. I'm confused lol.
This may not be the right place for such an elementary question, if so hopefully the mods will move it.

Homework Helper
What equation, given the mass of ball, the black hole, and initial distance, could yield the ball's velocity at a given distance in its descent?

force acting on it is "Gmm/r^2"

But how fast "r" changes depends on the acceleration of the object. I

As I understand your difficulty it is that you want velocity as a function of distance. You have force (and, thereby, acceleration) as a function of distance. You could integrate acceleration over time to get velocity, but you only have acceleration as a function of distance, not as a function of time.

The easy approach is to focus on energy instead of velocity. You have force as a function of distance. Force times incremental distance yields work. If you integrate force over distance, you get total work done. Which tells you energy. And that gives you velocity.

This approach is how one derives the potential energy associated with position in an unchanging gravitational field. The resulting formula for gravitational potential energy might or might not be familiar to you already.

• Nugatory and PeroK
Staff Emeritus
I haven't taken any physics outside of high school. I do know that the acceleration of the object is it's force over its mass. And the force acting on it is "Gmm/r^2". I guess I just need to know how to find its acceleration as "r" gets lower? But how fast "r" changes depends on the acceleration of the object. I'm confused lol.
This may not be the right place for such an elementary question, if so hopefully the mods will move it.

The force is not a constant in this case as the object moves closer towards the big object. So to be able to solve this, you will need calculus, assuming you want to approach this via kinematics. Are you equipped with this math to do that? It is useless to proceed with this if you don't.

Secondly, do you want to assume that the bigger mass is significantly bigger, and that it stays fixed in space as the other one moves?

BTW, I am always weary whenever a (new) member posts something and claiming it to be "simple". It seldom is.

Zz.

Why are you guys making it needlessly complicated? Just write conservation of mechanical energy in orbit around a massive body. That's as simple as it gets, no calculus involved.

@alexchamp29 A body under free fall in a central gravitational field will have some associated initial potential energy (by convention 0 if infinitely far away) and kinetic energy (0 if stationary). After a while of falling towards the central object, some of the potential energy will have transformed into kinetic energy, where the sum of both kinds of energies is the same before (subscript 'i') and after ('f').
I.e.:
$$E_{ki}+E_{pi}=E_{kf}+E_{pf}$$
$$\frac{mV_i^2}{2}+(-\frac{GMm}{r_i})=const=\frac{mV_f^2}{2}+(-\frac{GMm}{r_f})$$
You'll need the mass of the central object (M), the initial velocity (0 if it's stationary), and both the initial and final distance (r) from the central object. The mass of the falling body itself (m) cancels out.

Note the expression for potential energy is different from the (likely) more familiar ##mgh##. That's because it takes into account the acceleration changing with distance.

If you look at the right side of the equation, you should see that reduction of final radius (i.e. getting closer to the massive body) makes the expression for potential energy more negative. This means that for the sum to remain constant, the kinetic energy must increase, i.e. final velocity must go up.

• m4r35n357, Nugatory and PeroK
Staff Emeritus
That's why I said if the OP wants to do it kinematically. jbriggs has already addressed the problem via the energy equation.

Zz.

Gold Member
2022 Award
Warning: I'm a freshman chemistry student. My math skills are elementary at best.

Image a ball in deep space. A baseball sounds nice.
Assume this ball had no initial velocity but suddenly finds itself caught in the sphere of influence of a massive body. A black hole sounds fun. The ball begins accelerating towards the black hole. What equation, given the mass of ball, the black hole, and initial distance, could yield the ball's velocity at a given distance in its descent?
Just so you know, the bolded is a meaningless statement. Velocity is frame dependent so you HAVE to say what it is stationary relative to.

For your scenario what you want to say is that the ball is not orbiting the BH and is being held away from it by some force, thus maintaining a constant (but not orbital) distance. It is then released and THAT'S where your question has meaning.

Mentor
A black hole sounds fun.
It is, but also much harder to solve except when the ball is still an appreciable distance away.

Until and unless you have a solid understanding of how to analyze the simplest form of this problem (central mass is so much heavier than the ball that it can be considered immobile, central mass is something like an ordinary planet so we don't need to mess with relativistic corrections to Newton's ##F=GMm/r^2##) it's probably best to focus on that case. That's what the answers above are doing... pay partcular attention to the posts by @Bandersnatch and @jbriggs444 but don't expect to be able to take on a black hole that way.

• jbriggs444
Gold Member
While it's worthwhile working this out for yourself, the following could be used as a check. Wheeler and Turner would refer to an object falling from rest at a specific r as the 'drip' frame (an object falling from rest at infinity would be rain frame and an object hurled inward at speed from a great distance would be hail frame). The equation for the velocity at any given r in the drip frame for a static black hole is-

$$v_{\text{shell}}=\left(1-\frac{2M}{r_o}\right)^{-1/2}\left(\frac{2M}{r}-\frac{2M}{r_o}\right)^{1/2}$$
where $v_{\text{shell}}$ is the velocity at a specific radius ($r$), $r_o$ is the radius the object was dropped from and $M$ is the geometric unit of mass for the central object where $M=Gm/c^2$. In the case of a bh, the mass of the infalling object can be dismissed, especially something like a baseball. While this gives the proper (i.e. local) velocity of the infalling object, if viewed from a great distance the velocity would be different (i.e. slower), to obtain this, multiply the answer by $(1-2M/r)$.

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