Need conversion and specific heat?

AI Thread Summary
The discussion revolves around calculating the conversion of kinetic energy from an asteroid impact into thermal energy and its effect on Earth's atmosphere. The user seeks to understand how to directly convert kinetic energy to thermal energy, primarily through friction, and how much energy is required to raise the entire atmosphere's temperature by one degree. The orbital characteristics of the asteroid, specifically 951 Gaspara, are provided to illustrate its potential impact, though it does not cross Earth's orbit. The conversation highlights the complexities of energy transfer during such impacts, including atmospheric expansion and energy dissipation. The user is conducting research on the potential climate effects of asteroid debris, aiming to quantify how much energy could be absorbed by the atmosphere.
forumasker
Messages
7
Reaction score
0
It's not really homework its more of my own thing, but I don't know where else I'd put it

Homework Statement

category Main belt (Flora family)
Orbital characteristics
Epoch 6 March 2006 (JD 2453800.5)
Aphelion 2.594 AU (388.102 Gm)
Perihelion 1.825 AU (272.985 Gm)
Semi-major axis 2.210 AU (330.544 Gm)
Eccentricity 0.174
Orbital period 3.28 a (1199.647 d)
Average orbital speed 19.88 km/s
Mean anomaly 53.057°
Inclination 4.102°
Longitude of ascending node 253.2lllll18°
Argument of perihelion 129.532°
Proper orbital elements
Physical characteristics
Dimensions 18.2×10.5×8.9 km [1]
Mean radius 6.1 km[2]
Mass 2–3×10^16 kg (estimate)
Mean density ~2.7 g/cm³ (estimate) [3]
Equatorial surface gravity ~0.002 m/s² (estimate)
Escape velocity ~0.006 km/s (estimate)
Rotation period 0.293 d (7.042 h) [4]
Albedo 0.22 [5]
Temperature ~181 K
max: 281 K (+8°C)
Spectral type S
Absolute magnitude (H) 11.46

1199.647*24 because 24 hours, then times 60 because 60 minutes in an hour,
then times another 60 because 60 seconds in a minute, and we get about
1.03*10^8 seconds.

19.88km/s-0km/s)/(1.03*10^8s-0s)= approximately 1.93^-7km/s/s

E(sub k) = 1/2mv^2

Kinetic energy = (1/2)(2.5*10^16kg)(19.88km/s)

Homework Equations



E(sub k) = 1/2mv^2
(v2-v1)/(t2-t1)
and the other stuff I'm looking for

The Attempt at a Solution



The solution is kind of what I'm asking for, I need a way to convert kinetic energy directly into thermal energy, and then I also need to find how much energy it takes to raise the entire Earth's atmosphere by 1 degree of something or 1 kelvin, preferably degree.
 
Physics news on Phys.org
forumasker said:
It's not really homework its more of my own thing, but I don't know where else I'd put it

Homework Statement




category Main belt (Flora family)
Orbital characteristics
Epoch 6 March 2006 (JD 2453800.5)
Aphelion 2.594 AU (388.102 Gm)
Perihelion 1.825 AU (272.985 Gm)
Semi-major axis 2.210 AU (330.544 Gm)
Eccentricity 0.174
Orbital period 3.28 a (1199.647 d)
Average orbital speed 19.88 km/s
Mean anomaly 53.057°
Inclination 4.102°
Longitude of ascending node 253.2lllll18°
Argument of perihelion 129.532°
Proper orbital elements
Physical characteristics
Dimensions 18.2×10.5×8.9 km [1]
Mean radius 6.1 km[2]
Mass 2–3×10^16 kg (estimate)
Mean density ~2.7 g/cm³ (estimate) [3]
Equatorial surface gravity ~0.002 m/s² (estimate)
Escape velocity ~0.006 km/s (estimate)
Rotation period 0.293 d (7.042 h) [4]
Albedo 0.22 [5]
Temperature ~181 K
max: 281 K (+8°C)
Spectral type S
Absolute magnitude (H) 11.46

1199.647*24 because 24 hours, then times 60 because 60 minutes in an hour,
then times another 60 because 60 seconds in a minute, and we get about
1.03*10^8 seconds.

19.88km/s-0km/s)/(1.03*10^8s-0s)= approximately 1.93^-7km/s/s

E(sub k) = 1/2mv^2

Kinetic energy = (1/2)(2.5*10^16kg)(19.88km/s)

Homework Equations



E(sub k) = 1/2mv^2
(v2-v1)/(t2-t1)
and the other stuff I'm looking for


The Attempt at a Solution



The solution is kind of what I'm asking for, I need a way to convert kinetic energy directly into thermal energy, and then I also need to find how much energy it takes to raise the entire Earth's atmosphere by 1 degree of something or 1 kelvin, preferably degree.

Hi Forumasker, Welcome to Physics Forums.

It's not entirely clear what your problem statement is, but I can surmise that you're trying to determine how much heat energy would be end up in Earth's atmosphere due to an impact by a asteroid or comet. Is that true?

Rigorous analysis would be difficult because there are several potential paths for energy to take from such a collision depending upon its nature; Depending upon the size, composition, and initial speed of the body it may wholly or partially disintegrate in the atmosphere due to friction. A big bow-shock pressure wave in front of the body also stirs up the atmosphere mechanically. If a large portion of the body makes it to the surface and undergoes "lithobraking", then its KE is dumped into the Earths surface via a cratering event. Some portion of that energy will still make it into the atmosphere as the "hot spot" cools.

The body for which you've provided data doesn't cross the Earth's orbit: it has a perihelion of 1.825 AU compared to the Earth's 1 AU nearly circular orbit. Perhaps you're just looking for sample characteristics for estimation purposes?

Supposing that the body was somehow perturbed into crossing the Earth's path, it's orbital velocity would be a function of its position in its orbit (distance from the Sun). You can work out its speed for any distance from the Sun using the concept of conservation of total mechanical energy. The total mechanical energy held by anybody in orbit around the Sun is
inversely proportional to the length of the body's major axis: ##\xi = -\mu/(2a)##.

If the body is "captured" by the Earth then it will gain additional energy from gravitational potential energy converted to kinetic energy as it falls into the Earth's local gravitational well. Whether or not this will be a significant contribution will depend upon the initial speed of the body w.r.t. the Earth -- is the body approaching the Earth "from behind" (prograde orbit) or from "in front" (retrograde orbit) or perhaps its trajectory is nearly perpendicular to the plane of the Earth's orbit? Maybe you're just looking for the maximum approach speed?

You should be able to search the web for values for the specific heat of air, as well as total mass estimates and volume of the atmosphere.

The atmosphere is in a container whose "lid" is provided by gravity. That means when energy is added it will not only go into raising its temperature but also expanding the atmosphere against its "lid" -- The atmospheric volume will get larger. So adding energy to the atmosphere is not quite the same thing as adding energy to air sealed in a fixed size container. For a given "dose" of energy, temperature changes will be moderated by some energy going into gravitational potential energy as the atmosphere expands.
 
Thank you, but I'm already aware of all of the factors, this a hypothetical. Answers would be much appreciated, such as how to directly convert kinetic energy to thermal energy, which I suppose is mostly through friction, as well as finding out how much energy it takes to heat Earth's atmosphere 1 degree. This is mostly for my research paper which I predict I will have to do at the end of the year but don't know for sure if I will have on theoretical physics and research, and I'm trying to see if asteroids/meteoroids could possibly account for any climate change, and the total amount of asteroid debris that has hit/entered the Earth in the past few years may be well over or under 10^16kg, but the asteroid I'm using is sort of a reference to see what happens if we can actually see what all that debris does in one single moment. 951 Gaspara just to see if all that energy was converted into thermal energy, since most asteroids that enter Earth's atmosphere aren't asteroids at all, they are meteoroids which are grains of dust and completely burn up upon entry, and because of that, most or at least 99.99 percent of a meteoroid's kinetic energy is converted into thermal energy via friction. So I'm just seeing "what happens if the amount of mass in Gaspara just "burned up" upon entry.
And then, I need to I guess use specific heat or some way see how much Earth's atmosphere heat's up when it absorbs all of that thermal energy, which may or may not be a lot, so I need someway to see how much all that energy raises the temperature of Earth's atmosphere.
I mean Hurricane Katrina contained over 180 atomic bombs of energy, so this theory isn't really looking good right now.
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top