Decending Lunar lander velocity

  • Thread starter Thread starter Susanne217
  • Start date Start date
  • Tags Tags
    Lunar Velocity
AI Thread Summary
The discussion revolves around calculating the final velocity of the Apollo Lunar Lander during its descent to the moon. The initial conditions include a mass of 200 kg, an initial velocity of 6 m/s, and engine thrust applied at specific intervals. A participant attempts to calculate the final velocity at t = 5 seconds using free-fall equations, leading to a negative velocity, which raises questions about the impact of thrust. Responses emphasize the necessity of incorporating the thrust force into the calculations and clarify that the acceleration must account for both gravity and engine thrust. The conversation highlights the complexity of applying standard physics formulas to real-world scenarios, particularly in understanding net forces and resulting accelerations.
Susanne217
Messages
311
Reaction score
0

Homework Statement



Lets imagine we are back in 1969 and the Apollo Lunarlander is on approach to the moon. It has a mass of 200 kg and initial velocity of v1 = 6 m/s. The Astrofolks push the big red button in the cabin and engine burst of energy to break the spacecraft where time = 2 sec the resulting Force of the engine is 800 Newton and time = 4 sec the Force produced of the engine is 800 Newton. Whats the velocity of the decending lunar lander at t = 5 sec?

My professor did a fancy calculation where he included the mass of spacecraft and the resulting force of the engine bursts. But I tried something else :)

Lets imagine that after the they used the engine to break the spacecraft that its free-falling towards to the moon then.

The Attempt at a Solution



v_{final} = v_{initial} - g_{moon}\cdot t where t = 5 sec and g_{moon} = 1.62 m/s^2

Then I plug this into the formula I get:

v_{final} = 6 m/s - 1.62 \cdot 5 = - 2.1 m/s

Since the engine isn't given bursts at t = 5 then can't the LL be see as free falling and we can disregard its mass? And since the LL is decending the the final velocity is negativ?

Sincerely
Susanne
 
Physics news on Phys.org
Your calculation would be correct, if the speed in question is the vertical speed, with positive up, and if there were no thrusting taking place between t = 0 and t = 5.

But, even though I have trouble understand the exact details in the problem text as you have quoted it, it do read like there is thrusting taking place and this will make the resulting speed come out different. No matter what method you use it must include the effect of the specified thrusting somehow.
 
filiplarsen said:
Your calculation would be correct, if the speed in question is the vertical speed, with positive up, and if there were no thrusting taking place between t = 0 and t = 5.

But, even though I have trouble understand the exact details in the problem text as you have quoted it, it do read like there is thrusting taking place and this will make the resulting speed come out different. No matter what method you use it must include the effect of the specified thrusting somehow.

The only equation I can come up with

W_1 + W_2 - m\cdot g = 1/2 \cdot m \cdot v^2

where W_1 and W_2 is the thrusting of the engine at t = 2 and t = 4 in KWh.

But if I plug this into this formula I don't get the right result

which is v = 2,1 m/s

What am I doing wrong?

Sincerely
Susanne
 
You cannot add a work and a force, they are of different units.
You have an initial speed.
You have two forces acting on the spaceship: gravity and engine thrust.
So what is the acceleration?
 
magwas said:
You cannot add a work and a force, they are of different units.
You have an initial speed.
You have two forces acting on the spaceship: gravity and engine thrust.
So what is the acceleration?

I found another interesting formula

Which states that if object of mass m is effected by a Net Force T then if the object moves vertically

then

\sum F_y = T = m \cdot a_y which is implies a_y = T/ m

Why is that these physics problems aren't formulated so one can use the off the shelf formulas in the book :(
 
Last edited:
Susanne217 said:
I found another interesting formula

Which states that if object of mass m is effected by a Net Force T then if the object moves vertically

then

\sum F_y = T = m \cdot a_y which is implies a_y = T/ m

Why is that these physics problems aren't formulated so one can use the off the shelf formulas in the book :(

Okay. You have the acceleration. Now you can come up with the speeds in different points of time.
You can use off the self formulas in the book, just have to know which ones and in what order:)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top