Question

A Lagrangian Point or gravitational dead zone is found between two stars, at a distance RA = 4020 AU from star A and RB = 20100 AU from star B where 1 AU = 149.6 million km. If the mass of star A is 6.8E43 kg, determine the force exerted on star A by star B. Derive and express algebraic solution in terms of givens: Ra Rb, ma and G.

Answer 1

The gravitational force is given by Newton's Law:

$F=G\frac{mM}{r^2}$

We know that in a lagrangian point the foreces cancel out, then in this case we have:

$\begin{gathered} -G\frac{mM_A}{r^2_A}+G\frac{mM_B}{r^2_B}=0 \\ -\frac{M_A}{r^2_A}+\frac{M_B}{r^2_B}=0 \end{gathered}$

From which we can find the mass of star B:

$\begin{gathered} -\frac{M_A}{r^2_A}+\frac{M_B}{r^2_B}=0 \\ \frac{M_B}{r^2_B}=\frac{M_A}{r^2_A} \\ M_B=\frac{r^2_B}{r^2_A}M_A \\ M_B=(\frac{20100}{4020})^2(6.8\times10^{43}) \\ M_B=1.7\times10^{45} \end{gathered}$

Now, that we know the mass of star B we can calculate the force exerted on star A by star B:

$F=G\frac{M_AM_B}{(r_A+r_B)^2}$

plugging the values given we have:

$\begin{gathered} F=G\frac{M_AM_B}{(r_A+r_B)^2} \\ F=6.67\times10^{-11}\cdot\frac{(6.8\times10^{43}\cdot1.7\times^{45})}{\lbrack(24120)(1.496\times10^{11})\rbrack^2} \\ F=5.92\times10^{47} \end{gathered}$

Therefore, the force exerted on star A by star B is:

$F=5.92\times10^{47}\text{ N}$