2009-11-20, 07:14 PM
(This post was last modified: 2009-11-20, 07:16 PM by KajitiSouls.)
We needs constants to do that problem =O
Mearth = 5.97x10^24 kg (I've seen 5.98 in some books)
Msun = 1.99x10^30 kg
r = 6,371.0 km = 6.371x10^6 m
R = 1.50x10^6 km = 1.50x10^9 m
G = 6.674x10^(-11) N(m/kg)^2
Potential and Kinetic energy formulas:
KEtranslational = 1/2mv^2
KErotational = 1/2Iω^2, where I = mr^2 for masses distributed at distance r (read: point masses) on a plane, and I = 2/5mr^2 for spheres.
PEgravitational = -GM[SIZE="1"]1[/SIZE]M[SIZE="1"]2[/SIZE]/R
For the record, 2.7*10^33 is WAY too small!
I'm not sure how your teacher/book wants it done, but the rotational kinetic energy of earth orbiting the sun beats your answer by far. There's also the gravitational energy to consider and the rotation of the earth itself.
Mearth = 5.97x10^24 kg (I've seen 5.98 in some books)
Msun = 1.99x10^30 kg
r = 6,371.0 km = 6.371x10^6 m
R = 1.50x10^6 km = 1.50x10^9 m
G = 6.674x10^(-11) N(m/kg)^2
Potential and Kinetic energy formulas:
KEtranslational = 1/2mv^2
KErotational = 1/2Iω^2, where I = mr^2 for masses distributed at distance r (read: point masses) on a plane, and I = 2/5mr^2 for spheres.
PEgravitational = -GM[SIZE="1"]1[/SIZE]M[SIZE="1"]2[/SIZE]/R
For the record, 2.7*10^33 is WAY too small!
I'm not sure how your teacher/book wants it done, but the rotational kinetic energy of earth orbiting the sun beats your answer by far. There's also the gravitational energy to consider and the rotation of the earth itself.