# Math20512

### FAQ

You are expected to know the following expressions for potential energy.

First two facts:

• By definition $$F = - \nabla V$$, which in 1 dimension is just $$F=-dV/dx$$, or $$V=-\textstyle\int F \,dx$$. Notice it follows that any force field in 1 dimension is conservative (friction is not, but it's not a field, as it depends on velocity).
• V is only defined up to a constant. If C is a constant, then V+C is as good as V (and there is no physical way to distinguish between the two).

## Springs

Hooke's law for a spring (or anything elastic) is that when it is extended through a distance x, the restoring force is proportional to the extension. The constant of proportionality is called the spring constant (or elastic constant) and is usually denoted k. Thus, $$F=-kx$$, and so by integrating, the potential energy is therefore $$\textstyle V=\frac12 kx^2$$.

## Gravity

(a) In the 'laboratory', the force on an object of mass m is F = -mg, where g is the acceleration due to gravity (and upwards is the positive direction). It follows that (by integrating), if the object is held at a height h above ground level, the gravitational potential energy is mgh.

(b) On the other hand, the gravitational force between two bodies follows an inverse square law (Newton's law of gravitation). If their masses are m1 and m2 and they are at a distance r apart, the force between them is $$F = -G m_1m_2/r^2$$ (in the direction along their line of centres), where G is the universal gravitational constant. It follows (by integrating) that the gravitational potential is $$V = \frac{Gm_1m_2}{r}$$.

#### Question

How are the two cases above related: gravity between two bodies and gravity in the lab? If you can answer that, then what exactly is the relationship between G and g?

Page last modified on Thu, 21 April 2011, at 08:22 (BST)
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