Question

(II) Two objects attract each other gravitationally with a force of 2.5 × 10 ^–10 N when they are 0.25 m apart. Their total mass is 4.00 kg. Find their individual masses.

Answer 1

Answer:

The individual masses of the two objects are 0,06 kg and 3,94 kg.

Explanation:

To solve this we need to use Newton's law of universal gravitation that states that the force of gravity between two massive bodies 1 and 2, is proportional to their masses M₁ and M₂ and inversely proportional to the square of the distance d that separates them:

                                        $F=\frac{G M_{1}M_{2}}{d^{2} }$

with $G=6,67 x 10^{-11} \frac{Nm^{2} }{kg^{2} }$ being the universal gravitational constant.

We are told:

1. the value of F when it says that the objects attract each other gravitationally with a force of $2.5x10^{-10}N$,
2. the value of d when it says that they are 0.25 m apart.

When it says that their total mass is 4.00 kg it means that:

                                        $M_{1} +M_{2} =4.00\ kg$

                                                 $M_{2} =4.00\ kg-M_{1}$

replacing all of this in the original equation $F=\frac{G M_{1}M_{2}}{d^{2} }$ we get

                                      $2.5x10^{-10}N=6.67x10^{-11}\frac{Nm^{2}}{kg^{2}} \frac{M_{1}(4.00kg-M_{1})}{(0.25)^{2}m^{2}}$

                                      $2.5x10^{-10}N=\frac{6.67x10^{-11}Nm^{2}}{(0.25)^{2}m^{2}kg^{2}}(M_{1}4.00kg-M_{1} ^{2})$

                                     $2.5x10^{-10}N=\frac{6.67x10^{-11}N}{(0.25)^{2}kg^{2}}(M_{1}4.00kg-M_{1} ^{2})$

                                     $\frac{2.5x10^{-10} }{6.67x10^{-11}}(0.25)^{2}kg^{2}=4.00kg\ M_{1}-M_{1} ^{2}$

if we put all terms in one side we get

                                     $M_{1} ^{2} -4.00kg\ M_{1} +\frac{2.5x10^{-10}}{6.67x10^{-11}}(0.25)^{2}kg^{2} =0$

                                      $M_{1} ^{2} -4.00kg\ M_{1} +0.23\ kg^{2} =0$

we have to solve a quadratic equation $ax^{2} +bx+c=0$ where a=1, b=-4.00 and c=0.23. Here x=M₁.

We replace these values in the quadratic formula for the roots:

                                      $x=\frac{-b+-\sqrt{b^{2}-4.a.c}}{2a}$

so we get

                                      $x=\frac{4.00+-\sqrt{16-4.1.0.23}}{2}$

                                      $x=\frac{4.00+-\sqrt{16-0.92}}{2}$            

                                      $x=\frac{4.00+-\sqrt{15.08}}{2}$

                                      $x=\frac{4.00+-3.88}{2}$

the possible roots are

                                      $x_{1} =\frac{4.00+3.88}{2}=\frac{7.88}{2}=3.94$

                                      $x_{2} =\frac{4.00-3.88}{2}=\frac{0.12}{2}=0.06$

given that M₁+M₂=4.00 kg the possible values of M₂ are then

                                      $M_{2}=4.00kg-3.94kg=0.06kg$

                                      $M_{2}=4.00kg-0.06kg=3.94kg$

Either one of the individual masses M₁ and M₂ could be 0.06 kg or 3.94 kg.