# Asteroid Collision Course and Sagging Backpack

#### LongTermStudent

We recently had a study group to look over some problems we were given in preparation for our upcoming exam. There were two problems that gave us a little bit of difficulty. There's a lot of stuff here, but I tried to trim it up as much as possible so as not to scare off any help. If I need to supply more information please let me know. Thanks in advance for any help/guidance provided.

1. The problem statement, all variables and given/known data
A NASA satellite has just observed an asteroid that is on a collision course with Earth. The asteroid has an estimated mass, based on its size, of 5 x 109 kg. It is approaching the Earth on a head-on course with a velocity of 600 m/s relative to the Earth and is now 5 x 106 km away.

Masteroid = 5 x 10^9 kg
Mearth = 5.98 x 10^24 kg
Vi = 600 m/s
ri = 5 x 10^9 m

a. With what speed will it hit the Earth’s surface?
b. How much thermal energy will be released?
c. How much will the impact change the orbital velocity of the Earth?

2. Relevant equations

Ki + Ui = Kf + Uf
½MAVi² - G(MEMA)/ri = ½MAVf² - G(MEMA)/rE
½mvf² - Ug = ½mv² - Ug
MAVA = (MA + ME)V
MAVA + MEVE = (MA + ME)Vf
Vf = (MAVA) / (MA + ME)

3. The attempt at a solution

Part A we worked out using the first two equations in our study group:

½ (5 x 10^9 kg)(600 m/s)² - G(5.98x10^24 kg)( 5 x 10^9 kg)/(5 x 10^9 m)

and we ended up with 1.12 x 10^4 m/s relative to Earth as our final answer.

Part B we were unsure how to work because we haven't covered anything in class regarding how much energy is released as thermal energy, light, and sound. Using the third and fourth equations from above we worked it out as such:

½(5 x 10^9 kg)(1.12 10^4 m/s)² - ½(5 x 10^9 kg)(5.017 x 10^-13)²

and ended up with a final answer of 3.136 x 10^17 J.

Part C also left me with a certain level of uncertainty. Using the last two equations listed, I ended up with 9.37 x 10^-12 m/s or 3.14 x 10^15 %

1. The problem statement, all variables and given/known data
Two trees are 7.6 m apart. Calculate the magnitude of the force F a backpacker must exert to hold a 16 kg backpack so that the rope sags at its midpoint by:

a. 1.5 m
b. 0.15 m

2. Relevant equations

Σt = (TsinΘ)(x) - mgh = 0

3. The attempt at a solution

For Part A we were confused as to what values we should use for x and h because the diagram shows the rope going across the trees to a pulley where the camper is applying the force to the rope. Some of us thought that we should use a distance of 8.18 m for x and 4.09 m for h based on the numbers we came up using trig. However, some argued that we should use distances of 7.6 m and 3.8 m because that is the distance between the trees. I maintained the opinion that the pulley would create more length in the rope, so I went with the first set of numbers I mentioned.

Θ = arctan(1.5/3.8) = 21.54º
Σt = Tsin(21.54º)(8.18 m) - (16 kg)(9.8 m/s²)(4.09 m) = 0

My final answer for part A was 213.54 N

Part B offered no confusion because the distances obtained using trig were essentially 7.6 m and 3.8 m

#### cristo

Staff Emeritus
I'll talk about question 2. I think there's a simpler way than using
Σt = (TsinΘ)(x) - mgh = 0, as we don't know h.

Try drawing a picture of the rucksack and the forces acting upon it. You know the weight acting downwards, and we wish to calculate the tension in the rope. (Note that, for the rucksack to be in equilibrium, the tension must be the same throughout the rope) Then try and resolve the tension into vertical components (remember the tensions of both sides of the rope acting on the rucksack) and you should be able to solve for T (the force the backpacker must apply to the rope)

EDIT: your answer is correct. I've just calculated it using the method i mentioned, and it gives the same answer!! Sorry!

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#### LongTermStudent

Thanks for checking that out. There were several opinions floating around our study group regarding that problem. Now it's really just part B of the first problem that's really bothering me. Is there a better equation to find the thermal energy released by this collision?

#### cristo

Staff Emeritus
Your answer is correct. I think the confusion is arising due to the definition of "thermal energy." The definition used here is that thermal energy is the sum of the internal potential, and kinetic energies of the system.

Using this definition, if you consider the asteroid just as it hits the earths surface then the total thermal energy of the system is the kinetic energy of the asteroid: T= (m v^2)/2 = 3.136 x 10^17 J. When the asteroid strikes the earth (presuming it stops at rest) all this energy will be released, and so the total thermal energy released will be the answer you give.

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#### MillerMakramall

hey guys, can you tell me how you did question number 2 ( the tree one) ?, if you can like post it here or send it to my e-mail, that would be great. millermakramalla@hotmail.com
thanks

#### cristo

Staff Emeritus
hey guys, can you tell me how you did question number 2 ( the tree one) ?, if you can like post it here or send it to my e-mail, that would be great. millermakramalla@hotmail.com
thanks
Wow, this is from ages ago! Anyway, we do not give answers out here, but rather provide tutorial assistance for the student. So, if you show what you've done thus far, then we will be able to help you.

#### MillerMakramall

oh ok, good to know.
but thanks anyway, i think i got the answer. you just divide it into triagnles and figure out the Tension, which should be the force needed

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