# Help: Projectile Motion with Forces Problem -- Lunar Lander

1. Oct 2, 2014

### That_EDGEy_KiD

1. I have a physics problem that I am having trouble understanding.

Here it is: Your lunar lander (with fuel and crew) has a mass of 14,700 kg. It is initially in a circular orbit around the Moon (i.e. Viy = 0 m/s) at an altitude of 30 km above the lunar surface. To land on the Moon, you first slow the orbital velocity and the lander begins to fall toward the lunar surface. The lander is designed for a surface impact speed of no greater than 1.00 m/s. The landing rocket on your lander produces 30,000 N of force. At what altitude above the lunar surface should you turn on the landing rocket so that you land safely? The Moon's gravity is 1.662 m/s2

I'm having trouble understanding where I need to begin as well as how I would go about solving this problem. I would appreciate any help or guidance.

2.
F=MA, ?

3. 30,000= 14,700 x a
= 2.041

Last edited by a moderator: Oct 2, 2014
2. Oct 2, 2014

### Staff: Mentor

Welcome to the PF.

It seems like one piece of information is missing from the problem definition. How much is the orbiter slowed to de-orbit? Does the problem say anything about the de-orbit maneuver?

And you should add in the equations for uniform circular motion (relating acceleration to velocity, etc.).

3. Oct 2, 2014

### That_EDGEy_KiD

From my updated understanding of the problem, what's basically going to happen is:
Neglecting the curvature of the moon the problem is basically this right?
1) The lander starts to fall starting at 30km more or less vertically towards the surface
2) He is going to pick up speed while doing this, this speed is described by v_1=gt_1, where g=1.66m/s²
3) At a certain point he is going to turn on his trusters and another acceleration will happen in the opposite direction. a=F/m = 30 000N/14.700kg = 2.04m/s².
I have to find what this point is so that his final velocity is 1m/s.
When he turns on his trusters, he already will have this initial downward velocity v_1 picked up by his fall, therefore the final velocity:
v_2=v_1-at. ''a'' in this case is equal to 2.04m/s² (upward) - 1.662m/s² (downward) = 0.38m/s²
so, v_2=v_1-0.38t_2
It needs to be 1m/s so
1=v_1-0.38t_2
Now using the another piece of information, the 30km; The first part of the fall, before the trusters turn on is described by:
x_1=1.662(t_1)²/2 (x axis pointed downards)
The second part of the fall:
x_2=v_1t_2 - 0.38(t_2)²/2 = (v_1)(t_2)- 0.19(t_2)²
Now I know that x_1 + x_2 = 30 000m
1.662(t_1)²/2+(v_1)(t_2)- 0.19(t_2)² = 30 000m (*)
From previous equations I know that t_1=v_1/g and t_2=(v_1-1)/0.38.
This means that I can plug this into (*) and have an expression in only v1.
After solving for v1, I get the number 17.02 m/s?
After doing this, how do I find the altitude where the thrusters were fist turned on?

4. Oct 2, 2014

### Staff: Mentor

Good work! What is the expression for the velocity as a function of time, given the acceleration...?

5. Oct 2, 2014

### That_EDGEy_KiD

Are we talking about the equation: v=v(initial) + a*t ?

6. Oct 2, 2014

Yes :-)

7. Oct 2, 2014

### That_EDGEy_KiD

I'm still confused on how I would find the altitude where the thrusters were fist turned on?

8. Oct 2, 2014

### That_EDGEy_KiD

Would I just solve for either x_1 or x_2 ?

9. Oct 2, 2014

### That_EDGEy_KiD

yea, that doesn't work