# Intuition behind gravitational potential

Gravitational potential at a point is equal to work done in bringing a unit mass from infinity to a particular point

That was the text book definition

$$V_{p} = -\frac{GM}{r}$$

If we calculate $$V_p$$ for the Earth surface with $$R_e = 6.4 \times 10^6 \text{ m}$$ and $$M_e = 6 \times 10^{24} \text{ kg}$$ we get

$$V_p = -6.35 \times 10^7 \text{ J/kg}$$

Q1) What does this number tell us? Does it mean I have to give a body of mass $$1 \text{ kg}$$ kinetic energy worth of $$6.35 \times 10^7 \text{ J}$$ in order for the body to leave this planet (escape Earth's gravitational influence) and after escaping Earth's influence its kinetic energy will be zero?

What does $$-ve$$ sign indicate?

Q2) Does higher magnitude of $$V_{p}$$ indicate higher potential energy of the body? If yes, we know that everything in this universe tries to reduce its potential so won't it make more sense if gravity doesn't attract but rather repel other mass as potential is zero at infinity, but it attracts? If higher magnitude of $$V_{p}$$ doesn't indicate higher potential energy of the body skip to next question.

Q3) We have two formulas to look at: $$U = mgh$$ and $$V_p = -\frac{GM}{r}$$. Let's deal with mass of $$1 \text{ kg}$$. Assuming that $$-ve$$ sign of gravitational potential has nothing to do with potential energy.

In the first formula we are multiplying by height, which means higher the body from the Earth surface higher is the potential which makes sense. Here $$U$$ and $$h$$ are directly proportional. The second formula doesn't make sense as we're dividing with height, $$V_p = -\frac{G M_e}{R_e+h}$$ here $$h$$ and potential are inversely related?

I'm confused regarding this, tell me where am I going wrong

The equations in consideration are: $$V = \frac{-GM}{r} \ = \frac{U_g}{m}$$ Here $$M$$ is the mass of the particle establishing the gravitational field. It follows that $$U_g = -\frac{GMm}{r}$$.

Q1) What does this number tell us?

The value you have obtained essentially means that $$-6.35 \times 10^7 J$$ of work needs to be done on an object of mass $$1\text{kg}$$ to bring it from $$r = ∞$$ to $$r = 0$$ at constant velocity.

The reason for it being negative is that the applied force would have to be directed opposite to the body's displacement. This is because the gravitational force is attractive and would otherwise cause the body to accelerate. Consequently, in order to maintain a constant velocity, a force must be applied in the opposite direction.

Q2) Does higher magnitude of Vp indicate higher potential energy of the body?

First, the gravitational potential energy is defined for the system of two bodies (or more), and not the body alone. The term "potential energy of the body" does not make sense.

As $$V = \frac{U_g}{m}$$ a higher value of $$V$$ would mean a higher value of $$U_g$$, the gravitational potential energy of the system.

so won't it make more sense if gravity doesn't attract but rather repel other mass as potential is zero at infinity, but it attracts?

Pay attention to the negative sign in the equation. As $$r \rightarrow 0$$ (as the force is attractive), $$V \rightarrow -∞$$, consistent with your claim of the potential decreasing.

...Here U and h are directly proportional. The second formula doesn't make sense as we're diving with height, $$V_p=−\frac{GMe}{Re+h}$$ here h and potential are inversely related?

Again, the negative sign plays a role here. $$V \propto -\frac{1}{r} \ \text{not} \ V \propto \frac{1}{r}$$

Hence, as $$r$$ increases, $$V$$ increases, consistent with the expression for the approximate formuala for the potential energy of a two mass system.

Hope this helps.

Q1) What does this number tell us? Does it mean I have to give a body of mass $$1 \text{ kg}$$ kinetic energy worth of $$6.35 \times 10^7 \text{ J}$$ in order for the body to leave this planet (escape Earth's gravitational influence) and after escaping Earth's influence its kinetic energy will be zero?

As with all potential energy functions, it is the change in the potential energy that matters. Negative change of the gravitational potential energy equals work done by the gravitational force

$$W = -\Delta U = -(U_2 - U_1)$$

Note that "change" $$\Delta$$ always means "final minus initial value". In other words, negative value that you see does not mean anything on its own, it is the difference that matters. You can add arbitrary offset to the gravitational potential energy, it will cancel out when you take the difference anyway.

The work done by the gravitational force as object moves from $$r$$ to $$\infty$$ is

$$W_g = -\Delta U_g = U_1 - U_2 = -G \frac{m_1 m_2}{r} + G \frac{m_1 m_2}{\infty} = -G \frac{m_1 m_2}{r}$$

The work is obviously negative because displacement and force are in the opposite direction. As for the interpretation of this result, from the work-energy theorem $$\Delta K = W$$ it follows that if you want the object to escape Earth and have zero final kinetic energy then you need to do

$$W = \Delta U_g - K_i = -W_g - K_i = G \frac{M_e m}{r} - \frac{1}{2} m v_i^2$$

amount of work. Assuming initial kinetic energy is zero, the number you got $$-6.35 \times 10^7 \text{ J/kg}$$ means that for every $$\text{kg}$$ of mass you need to do $$+6.35 \times 10^7 \text{ J}$$ of work for object to escape Earth's gravitational influence.

Q3) We have two formulas to look at: $$U = mgh$$ and $$V_p = -\frac{GM}{r}$$.

...

I'm confused regarding this, tell me where am I going wrong

The equation $$\Delta U = m g \Delta h$$ is only an approximation which works well close to the Earth (planet) surface where $$g$$ is assumed to be constant. But value of $$g$$ changes with height (distance), so the correct expression is $$U = - G \frac{m_1 m_2}{r}$$

$$\Delta U = U_2 - U_1 = -G m_1 m_2 \Bigl( \frac{1}{r + \Delta h} - \frac{1}{r} \Bigr) = -G \frac{m_1 m_2}{r^2} \frac{\Delta h}{1 + \Delta h/r}$$

Now if you take $$m_2 \equiv m$$, $$m_1 = M_e$$, $$r = R_e$$, and $$\Delta h \ll R_e$$, the term $$\Delta h / R_e$$ becomes very small, and the above equation becomes

$$\Delta U = - m G \frac{M_e}{R_e^2} \frac{\Delta h}{1 + \Delta h/R_e} \approx m g \Delta h$$

which explains why gravitational potential energy can be approximated as $$U = m g h$$ close to the Earth surface.