0
$\begingroup$

The force experienced by a particle of mass $m$ in a planet's gravitational field along the z-axis, where $G$ is a constant is $$F = -\frac{GmM}{z^2}$$ What is the potential energy corresponding to the equation using energy conservation?

So using a conservation of energy: $$E(t) = T(t) + V(t) = \frac{1}{2}mv^2 + mgh$$

We have acceleration vector: $$\vec{a}(t)=\begin{pmatrix}0 \\ -g\end{pmatrix}$$

thus we have a velocity vector : $$\vec{v}(t)=\begin{pmatrix}v_{0}\cos(\theta) \\ - gt\end{pmatrix}$$

integrating again we have a position vector : $$\vec{r}(t)=\begin{pmatrix}v_{0}\cos(\theta)\:t \\ -\frac{g}{2}t^{2}\end{pmatrix}$$

So potential energy of an equation becomes : $$V(t)=mgh=mgs_{y}(t)=mg\left(-\frac{g}{2}t^{2}\right)$$ Am I on the right track? Any help will be appreciated

$\endgroup$

migrated from math.stackexchange.com Oct 4 '16 at 19:43

This question came from our site for people studying math at any level and professionals in related fields.

  • $\begingroup$ All the formulas you're using only work in fields of constant gravity. Clearly this situation involves a field of varying gravity, so the formulas you're using don't apply. $\endgroup$ – Anon Oct 4 '16 at 14:39
  • $\begingroup$ any recommendations on what kinds of formulas I should be using? Thank you $\endgroup$ – Michael.K Oct 4 '16 at 14:40
1
$\begingroup$

Conservation of energy tells us that the change in the kinetic energy of an object is equal to the work done on the object:$$K_{final}-K_{initial}=W.$$

We also know that we can define a total mechanical energy of a system to be the sum of the kinetic energy plus terms we call potential energy: $$E=K+U,$$ but the potential energies are due to force interactions inside the system.

The energy of a system can be changed by doing work on the system, and only by doing work on the system. Energy is never spontaneously created or destroyed. That's the essence of conservation of energy.

If gravity does work on an object, the kinetic energy of that object will change. If we include the gravitational interaction in the definition of the system, then the system has potential energy, and we might be able to treat the system as having a constant total mechanical energy if no outside force acts on the system, and therefore no work is done on the system (note that this is a special case of conservation of energy--energy is always conserved, but system energy is not always constant): $$E_{initial}=E_{final}$$ $$K_i+U_i=K_f+U_f$$ $$U_i-U_f=K_f-K_i=W$$ $$U_f-U_i=-W$$

A potential energy function can be found by calculating the negative of the work done by a force on an object as the object moves from some reference point (where the potential energy is defined to be zero) to some other point in space. Work done by a force $\vec{F}$ is defined as $$\int\vec{F}\cdot d\vec{r}$$ along some path. If the work is independent of the path but only depends on the beginning and ending locations, a potential energy function exists: $$U(\vec{R})=-\int_{ref}^{\vec{R}}\vec{F}\cdot d\vec{r}.$$

In your case, you have a force which is acting in the negative z direction. The reference point will be (as standardized in physics) $z=+\infty$ and your final location should be some finite $Z$: $$U(Z)=-\int_{\infty}^{Z}\frac{-GmM}{z^2} dz.$$

You can do the integral yourself. Yes, the function is always negative.

$\endgroup$
0
$\begingroup$

The potential energy of a system is given by the formula .... integral of F.dz with negative sign where F is the external force on the system and therefore the potential energy of the system corresponding to the equation in the question will simply be equal to integral of F.dz which is integrated within given limits.

The answer will be GMm/z. The formula used indeed comes from energy conservation by which we can say that work done by gravitational force will be stored in the form of potential energy of the system and actually work done by a variable force is actually equal to Integral of F.dz which is of opposite sign when compared with potential energy.

$\endgroup$
  • $\begingroup$ I think it will be - of that as potential energy here is attractive in terms of physics $\endgroup$ – Archis Welankar Oct 4 '16 at 15:02
  • $\begingroup$ The force itself is negative and hence the potential energy stored will be positive as a postive force is repulsive in nature and vice versa...correct me if I am wrong.... $\endgroup$ – SirJMP Oct 4 '16 at 15:06
  • $\begingroup$ I was told by professor that it just indicates attractiveness of force its just a sign convention maybe i am wrong $\endgroup$ – Archis Welankar Oct 4 '16 at 17:00
  • $\begingroup$ It must be negative if the force is acting in the negative z direction. It's not merely a sign convention. The potential energy must increase as you move against the force. $\endgroup$ – Bill N Oct 4 '16 at 20:30

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.