PX155 - A7 - gravitational force

• two point masses, $m_1$ and $m_2$, separated by distance $r$
• ${\overrightarrow{F}}_{12}=$ force due to body 1 on body 2
• $\hat{r}=$ unit vector pointing from $m_1$ to $m_2$

newton's law of gravity

• attractive, so, $m_2$ is pulled towards $m_1$ in direction $-\hat{r}$
• $G=6.674\times {10}^{-11}N{m}^{2}k{g}^{-2}$

newton's shell theorem

• consider a spherical shell

• point mass outside the shell: force of point mass due to the total mass of the shell is the same as if all the mass of the shell were concentrated into a point at its centre

• point mass inside the shell: gravitational force due to the shell precisely cancels, hence it is zero

• a solid sphere can be considered a series of concentric shells with a common centre.

• outside of the sphere can be treated as a point mass for gravitational purposes

• eg: three identical iron spheres with radius $(R)=5m$ are placed in contact

  • to calculate: force on $A$ due to $B$ and $C$:

- $r=2R$, $m_A=m_B=\frac{4}{3}\pi r^3\rho$ : $$\vec F_{BA}=\frac{G\frac{16}{9}\pi^2R^6\rho^2}{4R^2}$$

- putting $R=5m$, $\rho =7874kg{m}^{-3}$ : $$\vec F_{BA}= \frac{4}{9} 6.674\times 10^{-11}\pi
{ #2}
\times5^4\times (7.874\times 10^3)^{-3}$$

• eg: find the weight a person with mass, $m=70kg$, loses by climbing the Eiffel Tower$$W_G=\frac{G{M}_{E}m}{R_E^2}$$$$W_T=\frac{G{M}_{E}m}{(R_E+h{)}^2}$$
  • $h=320m$, $R_E=6371km$, $M_E=5.972\times {10}^{2}4kg$
  $$W_G-W_T=G{M}_{E}m[\frac{(R_E+h{)}^2-R_E^2}{R_E^2(R_E+h{)}^2)}]$$

- since $2R_{E}h>>h^2$: $$W_G-W_{T} \approx GM_{E} m \left[\frac{2h}{R_{E}^{3} }\right] 2 \left( \frac{GM_{E}}{R_{E}^{2}} \right) m \left( \frac{h}{R_{E}} \right)$$