# Why we say that total energy should be zero at infinity in the derivation of escape velocity?

I am little confuse about escape velocity. I have been taught that potential energy is arbitrary and hence we always assume it zero at infinity but actually it can or cannot be zero. My question is while deriving escape velocity we say that at infinity total energy = Kinetic + potential =0 and hence we can escape the gravitational field, why we consider infinity to be that point from where we can escape, why not some other points?

At infinity potential energy is not zero as there is always an gravitational force acting on the body, so even if the kinetic energy gets zero the object due to potential energy will come back. Where i am going wrong?

The gravitational force is $$F=-\dfrac{GMm}{r^2}$$ which goes to zero as $$r\to\infty$$. As $$r$$ increases, the gravitational force decreases and eventially, at infinity, dies off.

Remember that the force is $$F = -\dfrac{dU}{dr}$$ where $$U$$ is the potential energy.

In nature, we are aware of the force, and to it we assign a calculated potential energy. Whereas differences in potential energy are measurable, the absolute potential energy is not and must be defined.

We proceed as follows, \begin{align} \Delta U &= - \int_{r_1}^{r_2} -\dfrac{GMm}{r^2} \, \mathrm dr \\ &= -\dfrac{GMm}r \bigg|_{r_1}^{r_2} \end{align}

We now want to define $$U(r)$$. What we could do is say $$r_2 = r$$, and let $$r_1\to\infty$$. This makes it so that $$U(r) = -\dfrac{GMm}r$$, which goes to zero as $$r$$ increases.

Another to way to think about this is $$\displaystyle U = \int \dfrac{GMm}{r^2} \, \mathrm dr=-\dfrac{GMm}r +C$$ where we are free to choose $$C$$ and set it to zero for convenience.

• The +C term is already present in the indefinite integral and need not be added as an extra term to get arbitrary absolute U(r). $\int \dfrac{1}{x^2} \, \mathrm dx=-\dfrac{1}x +C$ (It disappears from the definite integral by subtracting $C-C$.)
– g s
Jun 11 at 18:47

Your mistake is claiming that there is a gravitational force at infinity, when in fact, it is precisely zero. You can see that from the limit

$$\lim_{r\rightarrow \infty} G\frac{Mm}{r^2} = 0$$

So an object at infinity will remain at infinity and will not be attracted back due to the gravitational force. That's why we define infinity to be the point where an object has escaped a gravitational source.

The above analysis is at the level of force. If you analyze at the level of potential energy, a key point is that the absolute value of potential energy doesn't matter; it is the difference in potential energy that matters. It is out of convenience that we define the potential energy to be zero at infinity.

• If we dont assume potential energy be zero at infinity than, total mechanical energy at infinity will not be zero, let say we assue earth surface to be our reference point, then at infinity the kinetic energy will be zero but not the potential energy Jun 11 at 16:32
• @MdFaiyaz Yes, but if we change PE at infinity to some non-zero value then we change PE at the Earth’s surface by the same value. So the difference in PE remains the same. And this difference is the KE required to reach infinity. Jun 11 at 17:35

Suppose that you don't want to use infinity as the zero of the potential energy $$U$$. Ok, that is fine since only energy differences matter. So, go ahead and use the expression: $$U = -\dfrac{GMm}r +C\;,$$ where $$C$$ is an arbitrary constant.

Note that as $$r\to\infty$$, we have $$U\to U_{\infty}$$ where $$U_\infty = -0 + C = C$$

You do have to use this expression for $$U$$ consistently. You can't use a different $$U$$ on different sides of an energy balance equation.

Anyways, balancing the total energy, say at the surface of earth (at radial distance $$R$$ from the center) and at infinity, gives: $$KE_{earth} + U _{earth} = KE_{\infty} + U_{\infty}$$ Or, setting $$KE_{\infty}$$ to zero, since we want to know the escape velocity $$v_e$$: $$\frac{1}{2}mv_e^2 -\frac{GMm}{R} + C= 0 - 0 + C\;.$$

So, regardless of what $$C$$ you choose, the constant cancels and you find: $$\frac{1}{2}mv_e^2 = \frac{GMm}{R}$$

Or $$v_e = \sqrt{\frac{2GM}{R}}$$

• Thank you for the answer but please tell me why we use infinity as the point to where we can escape the gravitational field, why not other point? Jun 12 at 8:38
• In this case "infinity" means "infinitely far away from the mass that is the source of the gravitational field from which we are trying to escape." The meaning of "escape velocity" is the velocity at which we can just barely get to infinitely far from the gravitational source (located at $r=0$ in the coordinates we chose).
– hft
Jun 12 at 18:15