Question

How to find gravitational potential energy component?

Answer 1

Answer:

The gravitational potential function, also known as gravitational potential energy, is:

{\displaystyle U=-{\frac {GMm}{r}},}U=-{\frac {GMm}{r}},

The negative sign follows the convention that work is gained from a loss of potential energy.

Derivation

The gravitational force between two bodies of mass M and m separated by a distance r is given by Newton's law

{\displaystyle \mathbf {F} =-{\frac {GMm}{r^{2}}}\mathbf {\hat {r}} ,}{\displaystyle \mathbf {F} =-{\frac {GMm}{r^{2}}}\mathbf {\hat {r}} ,}

where {\displaystyle \mathbf {\hat {r}} }\mathbf {\hat {r}}  is a vector of length 1 pointing from M to m and G is the gravitational constant.

Let the mass m move at the velocity v then the work of gravity on this mass as it moves from position r(t1) to r(t2) is given by

{\displaystyle W=-\int _{\mathbf {r} (t_{1})}^{\mathbf {r} (t_{2})}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot d\mathbf {r} =-\int _{t_{1}}^{t_{2}}{\frac {GMm}{r^{3}}}\mathbf {r} \cdot \mathbf {v} \mathrm {d} t.} W=-\int^{\mathbf{r}(t_2)}_{\mathbf{r}(t_1)}\frac{GMm}{r^3}\mathbf{r}\cdot d\mathbf{r}=-\int^{t_2}_{t_1}\frac{GMm}{r^3}\mathbf{r}\cdot\mathbf{v}\mathrm{d}t.

The position and velocity of the mass m are given by

{\displaystyle \mathbf {r} =r\mathbf {e} _{r},\qquad \mathbf {v} ={\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t},}{\displaystyle \mathbf {r} =r\mathbf {e} _{r},\qquad \mathbf {v} ={\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t},}

where er and et are the radial and tangential unit vectors directed relative to the vector from M to m. Use this to simplify the formula for work of gravity to,

{\displaystyle W=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}(r\mathbf {e} _{r})\cdot ({\dot {r}}\mathbf {e} _{r}+r{\dot {\theta }}\mathbf {e} _{t})\mathrm {d} t=-\int _{t_{1}}^{t_{2}}{\frac {GmM}{r^{3}}}r{\dot {r}}\mathrm {d} t={\frac {GMm}{r(t_{2})}}-{\frac {GMm}{r(t_{1})}}.} W=-\int^{t_2}_{t_1}\frac{GmM}{r^3}(r\mathbf{e}_r)\cdot(\dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_t)\mathrm{d}t = -\int^{t_2}_{t_1}\frac{GmM}{r^3}r\dot{r}\mathrm{d}t = \frac{GMm}{r(t_2)}-\frac{GMm}{r(t_1)}.

This calculation uses the fact that

{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}r^{-1}=-r^{-2}{\dot {r}}=-{\frac {\dot {r}}{r^{2}}}.} \frac{\mathrm{d}}{\mathrm{d}t}r^{-1}=-r^{-2}\dot{r}=-\frac{\dot{r}}{r^2}.

Explanation: