- #1

forumasker

- 7

- 0

## 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.