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(mM)/(r^2)`

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

`\begin{gathered} -G(mM_A)/(r^2_A)+G(mM_B)/(r^2_B)=0 \\ -(M_A)/(r^2_A)+(M_B)/(r^2_B)=0 \end{gathered}`

From which we can find the mass of star B:

`\begin{gathered} -(M_A)/(r^2_A)+(M_B)/(r^2_B)=0 \\ (M_B)/(r^2_B)=(M_A)/(r^2_A) \\ M_B=(r^2_B)/(r^2_A)M_A \\ M_B=((20100)/(4020))^2(6.8*10^(43)) \\ M_B=1.7*10^(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(M_AM_B)/((r_A+r_B)^2)`

plugging the values given we have:

`\begin{gathered} F=G(M_AM_B)/((r_A+r_B)^2) \\ F=6.67*10^(-11)\cdot((6.8*10^(43)\cdot1.7*^(45)))/(\lbrack(24120)(1.496*10^(11))\rbrack^2) \\ F=5.92*10^(47) \end{gathered}`

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

`F=5.92*10^(47)\text{ N}`