Gravitational Fields

Newton's lawn of gravitation

The force between two point masses is proportional to the product of their masses and inversely proportional to the square of their separation

F_g = \frac{GMm}{r ^ 2}

$F_g$ Force due to gravitation (N)
$G$ Gravitational constant $6.67*10^{-11}$
$\text{ and } n$ Two pointer masses
$r^2$ Separation of C.O.Ms

$F_g = mg$ is related to this formula: $g = \frac{GM}{r^2}$

Gravitational field strength

The gravitational force exerted at a point per unit mass on a small test mass at that point

g = \frac{GM}{r ^ 2}

Gravitational potential energy

Recall $\Delta{E_p} = mg\Delta{h}$ but now it's $g = \frac{GM}{r^2}$

$h = r$

E_p = \frac{-GMm*h}{r^2} = \frac{-GMn}{r}

Where did the negative come from?

Gravitational potential energy is defined as the work done in bringing a small test mass from infinity to that point

Maximum $E_p$ is at $r = \infty$ so at any $r$ less than $\infty$ there must be a smaller $E_p$ . Since at $r = \infty$ , $E_p = 0$ , any smaller value must be negative

Gravitational potential

The gravitational potential at a point is work done per unit mass in bringing a small test mass from infinity to that point

\phi = \frac{-GM}{r}

Escape velocity

The velocity at which an object on the surface of a body/mass must be propelled to in order not to return to that body under their mutual gravitation attraction

i.e energy required to get $r = \infty$ m from earth

\Delta{E_p} (gain) = E_p (\infty) - E_p (surface)

= 0 - \frac{-GMm}{r_e}

\frac{-GM\cancel{m}}{r_e} = \frac{1}{2} \cancel{m} v^2

\sqrt{\frac{2GM}{r_e}} = v_{escape}