The zero-velocity value of the Jacobi integral at the L2 point of the Earth-Moon system is C= -1.59411. Suppose a spacecraft is newar Earth (r2=1/60 in our descriptions). Approximately waht speed must the craft have realtive to the rotating system if it is able to reach the L2 point? In considering the problem, you may ignore all terms in the expression for the Jacobi integral that are "small", which for the purposes of this problem we will take to mean that they are of the order of u (mew) or smaller. (Note that you must explain why they are small)

To solve this problem, we need to find the approximate speed the spacecraft must have relative to the rotating system in order to reach the L2 point of the Earth-Moon system. We are given the zero-velocity value of the Jacobi integral at the L2 point, which is C = -1.59411.

The Jacobi integral is a constant of motion in the Earth-Moon system and it is defined as:

C = 2(T - U) - Ω²R²

where T represents the kinetic energy, U represents the potential energy, Ω represents the angular rotation rate, and R represents the distance from the center of the system.

In this problem, we are asked to ignore terms in the Jacobi integral that are "small" or of the order of u (mew) or smaller. To explain why these terms are small, let's analyze the Earth-Moon system. Since we are considering the L2 point, which is a Lagrange point, the forces acting on the spacecraft are mainly gravitational forces.

In the expressions for the kinetic energy (T) and potential energy (U), the terms involving the gravitational forces will dominate while the terms involving other small forces, such as tidal forces, will be much smaller. Therefore, we can neglect these small forces and consider only the gravitational forces.

Now, let's focus on the gravitational potential energy term (U). The potential energy is given by the formula:

U = -GMm / r

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the spacecraft, and r is the distance between the spacecraft and the Earth.

In this problem, we are told that the spacecraft is near Earth, with r2 = 1/60, which means the distance between the spacecraft and the Earth is 1/60 times the distance between the Earth and the Moon. Since the spacecraft is near Earth, the mass of the Earth is much larger than the mass of the spacecraft (M >> m), so we can approximate the potential energy as:

U ≈ -GM / r

Now, let's consider the term involving the angular rotation rate (Ω) and the distance from the center of the system (R). The angular rotation rate Ω represents the rotation of the Earth-Moon system, and R represents the distance of the spacecraft from the center of mass of the system.

In this problem, we are assuming that the spacecraft is near the Earth, so its distance from the center of mass of the system (R) is much smaller than the distance between the Earth and the Moon. Therefore, we can neglect this term in the Jacobi integral as it would be of the order of u (mew) or smaller.

With these considerations, we can simplify the Jacobi integral to:

C = 2T - Ω²R²

Since we are near the Earth, we can approximate the kinetic energy (T) as:

T ≈ 1/2 Mv²

where v is the velocity of the spacecraft relative to the rotating system.

Now, substitute the approximations for T and U into the Jacobi integral equation:

C = 2(1/2 Mv²) + GM / r

C = Mv² + GM / r

Since we are looking for the approximate speed the spacecraft must have relative to the rotating system to reach the L2 point, and C is a constant value at the L2 point, we can set C equal to its value (-1.59411) and solve for v:

-1.59411 = Mv² + GM / r

To find the approximate speed (v), you would need to know the values of M (mass of the Earth), G (gravitational constant), and r (distance between the spacecraft and the Earth).