-1
$\begingroup$

I'm familiar with the potential energy equation, but I'm concerned with the value of 'g' in it. I know that, at sea level, earth's gravitational acceleration is 9.81 m/s/s. So I know that within the normal bounds of physics the minute change in the gravitational acceleration due to the change in height is negligible. I'm running some calculations where the change in gravity is major enough for it to make a difference, and I'm trying to calculate the maximum velocity a body will reach right before it stops falling, which the kinetic energy contained in this will equal the potential energy before the fall (assuming a stationary body at the beginning). So I needed to calculate gravity and substitute the acceleration for the 'g' in the potential energy equation, and to save time, I combined the equations to get:

$$P.E. = \frac{G M_1}{H}$$

where G is the gravitational constant, M is the mass of the object (in this case the planet) that the body is falling to, and H is the distance from the surface. In order for the calculation to be precise, the gravitational acceleration has to change as the height does, which is where calculus comes in. Because I'm working with accelerations and need to take the sum of all the accelerations between heights (to yield the final velocity of the fall), I thought to integrate, as it's basically area under a curve. But if you look at what the graphs are representing, it's a little weird. The initial equation marked acceleration over distance $\frac{m/s^2}{m}$ which is really just per second squared (no numerator, right?). So the integral should represent per meter second squared, if not acceleration per meter squared. These units don't seem to make sense. Am I going about accounting for the change in gravity wrongly, or is my calculus wrong? Ultimately what's the proper way to calculate potential energy of a body when the change in gravitational acceleration is not negligible? I know that the law of conservation of energy states that the kinetic energy at the end of the fall (technically right before) should equal the potential energy before the particle falls. However, I can't find a way to calculate the velocity, accounting for the change in acceleration that accompanies the change in height.

$\endgroup$
1
$\begingroup$

Actually, it is more correct to say that the change in kinetic energy over the course of the fall is equal to the (negative) change in the potential energy that occurred during the fall. The initial potential energy is $$ U_i = -\frac{GMm}{r_i} $$ and the final potential energy is $$ U_f = -\frac{GMm}{r_f} $$ where $U$ is the gravitational potential energy, $M$ is the mass of the earth, $m$ is the mass of whatever is falling, and $r$ is the distance from the center of the earth (note that $U\to 0$ as $r\to \infty$). In this case, if the falling object hits the ground, then $r_f$ will be the radius of the earth. Then, to find the speed of the object as it hits the ground, you can use conservation of energy, like you mentioned: $$\Delta T = -\Delta U$$ where $T$ is the kinetic energy. Assuming the object starts from rest, your $\Delta T$ just becomes $T_f$. And keep in mind that the change in a quantity is calculated as the difference between its final and initial values.

$\endgroup$
  • $\begingroup$ You said that as U approaches 0, r should too. Clarify that because it's an inverse relationship so as U approaches 0, r should approach infinity. Also, with this model, potential energy decreases as height (r) increases, which really shouldn't be. I agree that the change in potential energy should equal the magnitude of the change in kinetic energy, but the relationship you describe suggests negative energy (as the increase in potential energy as height decreases is acdounted for in a negative (imaginary) kinetic energy), which shouldn't directly apply here. $\endgroup$ – Jean Valjean Feb 15 '16 at 22:58
  • $\begingroup$ Sorry, that should have said $U\to 0$ as $r\to \infty$. As for the potential energy decreasing as height increases, note the minus sign in front of the expression for $U$. Say $U=-10$ J on the surface of the earth. And as $r\to\infty$, $U\to 0$, so it's increasing, not decreasing (i.e. $0>-10$). $\endgroup$ – Mike Bell Feb 15 '16 at 23:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.