Decending Lunar lander velocity

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Homework Help Overview

The discussion revolves around a physics problem involving the descent of a lunar lander, focusing on its velocity as it approaches the moon. The scenario includes initial conditions such as mass, initial velocity, and forces acting on the lander, specifically engine thrust and gravitational acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various methods to calculate the final velocity of the lunar lander, questioning the impact of thrust and gravity. Some participants attempt to apply kinematic equations while others express confusion over the problem's details and the correct application of forces.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning assumptions about thrust and free fall. Some guidance has been offered regarding the need to account for thrust in calculations, but no consensus has been reached on the correct approach or final velocity.

Contextual Notes

There are indications of confusion regarding the units of work and force, as well as the formulation of the problem, which may be affecting participants' ability to apply standard physics formulas effectively.

Susanne217
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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
 
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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:)
 

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