Lunar Crash Trajectory Problem

[mastash]

Homework Statement

As a command module, orbiting the moon passed through a point A (120km altitude), it released a lunar excursion module, which crashed into the moons surface at point B. Knowing the angle AOB was 50 deg (where O is the centre of the moon), determine the velocity of the lunar excursion module, when it was cast adrift, relative to the command module.

Homework Equations

Also know the point A is the apogee of the elliptic crash trajectory, and the mass of the moon is 0.0123 times the mass of the earth. Therefore mass of moon = 7.36 × 10^22 kg.

The Attempt at a Solution

This is my first real attempt at a orbital question so I am kind of winging it. Any help would be greatly appreciated.

 

[jbsteele]

I tried to convert the problem into a 2-body problem, using the mass of the command module as zero, and the moon as the other body. I then used the equation v^2 = GM/r to find the velocity of the lunar excursion module at point A. v^2 = (6.67 × 10^-11) * (7.36 × 10^22) / (1737.1 × 10^6 + 120 × 10^3)v^2 = 1.103 × 10^5 m^2/s^2v = 1046.9 m/sThis seems too low, so I must have done something wrong. Any help would be greatly appreciated.

 

[tomcue]

I would approach this problem by first considering the basic principles of orbital mechanics. The velocity of an object in orbit is determined by its altitude and the gravitational pull of the body it is orbiting. In this case, the lunar excursion module was released from the command module at a height of 120km above the moon's surface. This means that the initial velocity of the excursion module would have been equal to the orbital velocity at this altitude, which can be calculated using the formula v = √(GM/r), where G is the gravitational constant, M is the mass of the moon, and r is the distance from the center of the moon.

Using the given mass of the moon (7.36 × 10^22 kg) and the radius of the moon (1,737.1 km), we can calculate the gravitational pull of the moon at a distance of 120km above its surface. This comes out to be 1.542 m/s^2. Plugging this value into the formula for orbital velocity, we get a velocity of 1,640.2 m/s.

However, we also need to take into account the angle AOB, which is given as 50 degrees. This means that the actual velocity of the lunar excursion module would have been the component of the orbital velocity in the direction of point B, which is equal to v * sin(50). This gives us a final velocity of 1,258.5 m/s.

It is important to note that this is the velocity of the excursion module relative to the command module. To determine the absolute velocity of the excursion module, we would need to know the velocity of the command module at the time of release. If we assume that the command module was in a circular orbit at the time, then its velocity would be equal to the orbital velocity at a height of 120km, which we have already calculated to be 1,640.2 m/s. Therefore, the absolute velocity of the excursion module would be 1,258.5 m/s + 1,640.2 m/s = 2,898.7 m/s.

In conclusion, the velocity of the lunar excursion module, relative to the command module, when it was cast adrift, was approximately 1,258.5 m/s.