What is the energy of the combined body after the collision?

[ori]

when body that it's is 100kg is at 1000km above earth, he collides with another body that it's mass is 50kg that moves infornt of him. the bodies stick together and becoming to one big body that it's mass is 150kg
what's the enegry of the new body?
given data:
Re=6400km
R2=1000km
G=6.7*10^(-11)Nm^2/Kg^2
Me=6*10^24kg
the right answer is -7696*10^6 J
please give me idea how to solve it

 

[Locrian]

Hah took me a second to see what you could be doing wrong.

First, make sure you convert your radius to meters.

However, I kept getting the wrong answer. THe problem? Your radius for the Earth is 300km short! 6700km is a better estimate, and with that I got the right answer.

Just plug the values into the gravitational potential energy equation.

 

[ori]

Hah took me a second to see what you could be doing wrong.

First, make sure you convert your radius to meters.

However, I kept getting the wrong answer. THe problem? Your radius for the Earth is 300km short! 6700km is a better estimate, and with that I got the right answer.

Just plug the values into the gravitational potential energy equation.

i don't understand why it is just to assign the value to the potential equation??
i also don't get the right answer with assign the values (with 6400 or 6700 for Re) it with the equation. ( i get something pretty close though -7831.17 * 10^6)

 

[Locrian]

Use 6.67*10^-11 for G and I bet you get even closer. Where did you get these values you are using for your constants? Using the wrong constant values is a quick way to guarantee a wrong answer.

As for why just assigning values, I'm not sure I understand your question. Are you asking why you use that particular equation? Maybe it was wrong for me to suggest just "plug n chug"ing, but this question is so basic its hard to avoid.

The potential energy of the object will be found by integrating the force with respect to distance from the center of the Earth to the object's orbit radius. The equation you'll get by doing this is the commonly used gravitational potential energy equation.

I hope I've answered your question; if not I'll try again or someone else can step in and help :D

 

[ori]

Use 6.67*10^-11 for G and I bet you get even closer. Where did you get these values you are using for your constants? Using the wrong constant values is a quick way to guarantee a wrong answer.

As for why just assigning values, I'm not sure I understand your question. Are you asking why you use that particular equation? Maybe it was wrong for me to suggest just "plug n chug"ing, but this question is so basic its hard to avoid.

The potential energy of the object will be found by integrating the force with respect to distance from the center of the Earth to the object's orbit radius. The equation you'll get by doing this is the commonly used gravitational potential energy equation.

I hope I've answered your question; if not I'll try again or someone else can step in and help :D

but there's also kinetic energy
plus the distance from Earth is changed i think

 

[Locrian]

Very true. To be honest, I assumed your problem actually wanted the potential energy. Since we know nothing about what kind of orbit they are in or the velocity of either body before collision, I don't think we are provided enough information to describe the total energy of the bodies. What's more, just finding the potential energy comes up with the correct solution.

What class is this for? Unfortunately I find that early physics classes contain many poorly worded questions. Could this be one of them?

 

[ori]

look u need get the exact answer with the given data
i thought about:
both bodies at the same speed , coz they collided, what means they at the same high.
so i tried momentum conservation :
1)m1*V-m2*V=m3*V3

and while r1 is 6400+1000 km
i used the equations
2) V^2=GM/r1
3) V3^2=GM/r3
4) E3=-G*M*m3/(2*r3)

E3=
(3-->4)
= -m3/2 * V3^2=
(assigned V3 that i took from 1)
= -m3/2 * (m1-m2)^2*V^2/m3^2 =
assign V from 2 and div from denominator&nominator m3
= -(1/2)*(m1-m2)^2 * GM/(r1*m3)
=-(1/2)*50^2*GM/(7400*1000*150)
=452.7 * 10^6
not gutta..

the question is from basic course at an univercity

 

[Locrian]

Well, the answer I get considering only potential energy is -7796*10^6. This gives me a very small error.

On the other hand, the equations you used didn't make any sense. To find their tangential velocity requires using equations that determine orbital speeds; you just use their potential energy and assume it is entirely converted to kinetic velocity, which would have a vector towards the center of the earth. Not to mention if they had the same speed and orbital radius they probably wouldn't be colliding in the first place.

I still think you are a victim of a badly worded question. I get less than 2% error finding the gravitational potential energy. I doubt they could mean something else.

 

[ori]

Well, the answer I get considering only potential energy is -7796*10^6. This gives me a very small error.

On the other hand, the equations you used didn't make any sense. To find their tangential velocity requires using equations that determine orbital speeds; you just use their potential energy and assume it is entirely converted to kinetic velocity, which would have a vector towards the center of the earth. Not to mention if they had the same speed and orbital radius they probably wouldn't be colliding in the first place.

I still think you are a victim of a badly worded question. I get less than 2% error finding the gravitational potential energy. I doubt they could mean something else.

u r wrong
i got the right answer(exactly :) )
thank u anyway

(
100V-50V = 150U
===> u=v/3

E = -G*150*Me/7400*1000+0.5*150*(v/3)^2
v is 7370)

 

[Locrian]

That's very strange! So you assume that both have the same initial velocity, assume their vectors are exactly opposite and use the wrong radius for the Earth and get the right answer. I must admit, I'm stumped.

I'm glad you got it though. Best of luck in the rest of your class. Post any others that stump you, I'd love to see more. Who knows, maybe next time I can help!