**We’re covering:**

- Gravitational Potential
- Gravitational Potential Energy
- Change in Gravitational Potential Energy
- Total energy of a satellite
- Escape velocity

**Let’s go!
**

** What is Gravitational Potential?**Work done per unit mass in bringing a small mass from infinity to a certain point.

**It is denoted by φ (phi).**

φ = W/m

Since W = Fr

W/m = (GMm/r^{2}) x r/m

W/m = GM/r*

***However, by definition (see below), we take φ as always being a NEGATIVE value.**

So,

**φ = – GM/r**

** Why is it always negative?**There are a few ways to explain this:

By Definition |
We define infinity to be our starting point.
At infinity, F = 0, g = 0, φ = 0. To bring a mass from infinity to a point IN a gravitational field, we do NEGATIVE work |

By Calculation |
Work done is the integral of F_{g}.
∫ F φ = W/m |

By applying the Principle of Conservation of Energy |
As the test mass moves from infinity to a point within the gravitational field, it LOSES the potential to do work (i.e. it can do less work).
In this sense it LOSES potential energy, & gains kinetic energy. |

** What is Gravitational Potential Energy?**Work done in bringing a small mass from infinity to a certain point.

E_{p} = mφ

E_{p} = – GMm/r

**Remember**: this is the energy needed to bring a **mass m** from **infinity** to a **point** within a gravitational field.

** What is the work done to move an object between 2 points within a gravitational field?**Work done = Change in Gravitational Potential Energy

ΔE_{p} = E_{pA} – E_{pB}

If the change in distance is small enough such that change gravitational field strength is negligible, we can actually say:

**ΔE _{p} = mgh**

This is where that expression comes from!

** What is the total energy of an orbital satellite?**Kinetic Energy + Gravitational Potential Energy.

ΣE = E_{k} + E_{g
}ΣE = ½ mv^{2} + (- GMm/r)

Since mv^{2}/r = GMm/r^{2 }(F_{c} = F_{g}),

½ mv^{2} = GMm/2r

Plugging this in:

ΣE = GMm/2r + (- GMm/r)

**Σ****E = – GMm/2r**

** What is escape velocity?**The minimum velocity required to escape a gravitational field.

The “edge” of a gravitational field is where the field strength g approaches 0.

Thus, an object which has escaped does not accelerate towards M.

When an object is launched from the surface at the escape velocity,

it will decelerate due to F_{g} until it reaches a velocity of 0 ms^{-1 }at the edge of the gravitational field.

Since the object is launched from the surface,

- Initial E
_{g}= -GMm/R - Final E
_{g}= -GMm/**∞**= 0 - Gain in E
_{g}= GMm/R

Loss of E_{k} = Gain in E_{g
}½ mv_{e}^{2} = GMm/r

**v _{e} = √(2GM/R)**

OR

**v _{e} = √(2gR)**

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