Question

Assume that \(L\) is much larger than the radius of either planet. What is the position, \(x\), of the spacecraft (given as a function of \(L, M_{1}\), and \(M_{2}\) ) if the net force on the spacecraft is zero?A spaceship of mass \(m\) is located between two planets of masses \(M_{1}\) and \(M_{2}\); the distance between the two planets is \(L\), as shown in the figure.

Short answer

The spacecraft must be at the position \(x = \frac{M_2L^2}{2M_2L - (M_1 - M_2)L}\) for the net force acting on it to be zero . This position depends on the distance between the planets \(L\) and their masses \(M_1\) and \(M_2\).

Step-by-step solution

The spacecraft is being pulled towards each of the two planets due to the force of gravity. The force exerted by planet 1 on the spacecraft is \(F_1\) and that exerted by planet 2 is \(F_2\). We will use Newton's Universal Law of Gravitation to find the expressions for these forces. #step2# Calculate the forces using Newton's Universal Law of Gravitation

According to Newton's Universal Law of Gravitation, the gravitational force between two objects with masses \(m_1\) and \(m_2\) separated by a distance \(r\) is given by: \(F = G \frac{m_1 m_2}{r^2}\) where \(G\) is the gravitational constant. In our case, \(F_1 = G\frac{M_1 m}{(L-x)^2}\) and \(F_2 = G\frac{M_2 m}{x^2}\). #step3# Equate the forces to achieve equilibrium

For the net force on the spacecraft to be zero, the forces exerted by the two planets must cancel each other out. Therefore, we can write: \(F_1 = F_2\) Substitute the expressions for \(F_1\) and \(F_2\) from step 2: \(G\frac{M_1 m}{(L-x)^2} = G\frac{M_2 m}{x^2}\) #step4# Solve for the position \(x\)

We want to find an expression for \(x\), so we need to solve the equation from step 3 for \(x\). First, cancel the gravitational constant \(G\) and the mass of the spacecraft \(m\) from both sides: \(\frac{M_1}{(L-x)^2} = \frac{M_2}{x^2}\) Next, cross-multiply: \(M_1x^2 = M_2(L-x)^2\) Expand the right-hand side: \(M_1x^2 = M_2(L^2 - 2Lx + x^2)\) Rearrange the equation and factor out \(x^2\): \(x^2(M_1 - M_2) = 2M_2Lx - M_2L^2\) Finally, solve for \(x\): \(x = \frac{M_2L^2}{2M_2L - (M_1 - M_2)L}\) This is the expression for the position \(x\) of the spacecraft as a function of \(L\), \(M_1\), and \(M_2\) for which the net force on the spacecraft is zero.