# What happens for a spoon if you carry it to the top of Mount Everest? [duplicate]

$$U=mgh$$

$$h \to \infty$$

$$U \to ?$$

The potential of a 100 gram-spoon is approximately 1 j when it's 1 meter away from the Earth's surface. It's still a spoon if you carry it to the top of Mount Everest. Its potential energy will be more than 8000-fold, while it's still the initial spoon without any differences! Why there's not any difference?

Will the matter explode if you increase its height, $h$, (referring to the earth) to infinity?

• Will the matter explode No. . . Is there a serious question here? Also, note that potential energy is, more properly, $U=-\frac{GMm}{r}$ Work must be done to bring the spoon far away. Jan 14, 2015 at 22:05

No, the object won't explode.

The equation $U=mgh$ is an approximation of the change in gravitational potential energy when moving the mass from a reference point to some different height (moving parallel to the gravitational field). So, in your example, it's the change in energy relative to the planet surface that is 8000 increased.

The gravitational potential energy due to two massive objects (a spoon and a spherical planet) is, from a Newtonian model, $$U = -\frac{Gm_sm_p}{r}$$ where r is the separation distance of the masses, center to center. As $r\rightarrow \infty$, $U\rightarrow 0$. Notice that $U$ is actually negative and increases toward zero. Also, the potential energy belongs to the system of objects, not to any one object.

The value $g$ is actually called the gravitational field strength, and at the surface of a planet, the planet has a field of $$g = \frac{Gm_p}{R^2_p},\ downward$$ where $R_p$ is the radius of the planet. for small distance changes near the surface of the planet we approximate the change in potential energy by $mgh$.

The attraction between two bodies of masses $M_1$ respectively $M_2$ is given by the law

$F = G\frac {M_1M_{Earth}}{r^2}$

To find the gravitational potential difference from the surface of the Earth ($R_0$ = 6.4Km) up to the peak of Everest, $R$ = (6.4 + 11)Km, you integrate by $r$

(1) $\Delta U = -G M_1M_{Earth}\int_{Earth}^{Everest} r^{-2} dr= G M_1M_{Earth}(\frac {1}{R_0} - \frac{1}{R_0 + h})$.

For values of $h$ much smaller than the Earth radius,

(2) $\Delta U ≈ \frac {G M_1M_{Earth}}{R_0^2} h = M_1g\ h$

But if $h \to \infty$ we have to use the formula (1), and you see that the 2nd term vanishes.

Because gravitational force stops pulling objects together at $h \to \infty$, the value $E = \frac {G M_1M_{Earth}}{R_0}$ is known as gravitational binding energy.