Calculating Needed Thrust for a Moon Landing

AI Thread Summary
To calculate the needed thrust for a Moon landing, a landing craft with a mass of 12,000 kg and an initial downward velocity of 15 m/s must overcome lunar gravity, which is approximately 1.635 m/s². The required upward thrust must not only counteract gravity but also provide the necessary deceleration to reduce the craft's velocity to zero upon landing. The calculation shows that a net upward force is essential for deceleration, leading to a required upward acceleration of -0.703125 m/s². This results in a thrust of 8437.5 N, indicating that this force must be applied upward to achieve a safe landing. Understanding the dynamics of thrust and acceleration is crucial for successful lunar landings.
ashvuck101
Messages
26
Reaction score
0

Homework Statement



A landing craft of mass 12,000 kg prepares itself for a Moon landing. When at a
vertical distance 160m above the Moon’s surface it’s downwards velocity is 15 m/s.
A retrorocket is fired to give the craft an upward thrust to slow its speed to zero as it
meets the surface. Calculate the needed thrust to decelerate the craft its initial
downwards velocity to zero as it lands. The value for gravity on the Moon is g/6.
(Where g is 9.81m/s2).


Homework Equations



f=ma


The Attempt at a Solution


Gravity =9.81/6
=1.635

F (craft)= ma
=12 000. 1.635
=19620
Therefore the craft will need a a thrust of 19620 N upwards. Is that right or is that other information needed??
 
Physics news on Phys.org
No that is not right. That upward thrust of 19620N merely counteracts the gravitational force acting on the landing craft. As such, there will be no net force on the craft, and it will continue moving downwards at a constant velocity of 15m/s.

In order to decelerate the craft, you are going to need a net upwards force acting on the craft. Based on the information in the question, what is the magnitude of the upward acceleration that the craft must undergo?
 
ok...but i can't calculate an upward force without, Acceleration I am only given constant velocity?? which means there is no acceleration
 
There has to be an acceleration for the craft's velocity to reduce to 0 m/s from 15 m/s. I think you should start by calculating the acceleration required to bring the craft's velocity down by 15 m/s in the given distance.
 
ok i did that and got -0.703125 as acceleration

then i got a force of -8437.5 N

so the craft needs a downward force of 8437.5N to come to 0 m/s? is it downward or upward??
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top