Andrew forgot to change km/s to m/s; that adds a factor of 10^6 when squared to the answer, making it:

KEe = 2.7E33 J

(Actually ~ 2.65E33 J when you use the real numbers instead of rough estimates as he did…….)

….and since the question specifically asks for the energy due to the orbital motion only, gravitational PE and KE of rotation about its own axis are thereby excluded from the calculation of ‘mechanical energy’…….

Philip J

9 years ago

Andrew B lost 6 decimals by plugging kilometers/sec into the equation as meter/sec. It should be 10^33, not 10^27.

The total energy of an orbiting body is a constant which includes both gravitational potential and kinetic energy. As the body approaches perihelion, it loses potential and gains kinetic. However, zero potential is an arbitrary reference. The most commonly chosen zero reference points are infinity and the surface of the body being orbited, i.e., the sun.

Andrew B

9 years ago

The By “mechanical” I am assuming you mean kinetic.
KE = 1/2 mv2

m = mass
v = the velocity of the orbiting body
E = 1/2 mv2

The velocity of the earth around the sun varies but the average is 29.783 kilometers per second.
The mass of the earth is 5.98 Ã— 10^24 kg. (the 24 is a power of the ten)

E = 0.5(5.98 Ã— 1024 kg)(29.783)2 = 3 x 10^24 x 900 = 2.7 x 10^27 joules

Andrew forgot to change km/s to m/s; that adds a factor of 10^6 when squared to the answer, making it:

KEe = 2.7E33 J

(Actually ~ 2.65E33 J when you use the real numbers instead of rough estimates as he did…….)

….and since the question specifically asks for the energy due to the orbital motion only, gravitational PE and KE of rotation about its own axis are thereby excluded from the calculation of ‘mechanical energy’…….

Andrew B lost 6 decimals by plugging kilometers/sec into the equation as meter/sec. It should be 10^33, not 10^27.

The total energy of an orbiting body is a constant which includes both gravitational potential and kinetic energy. As the body approaches perihelion, it loses potential and gains kinetic. However, zero potential is an arbitrary reference. The most commonly chosen zero reference points are infinity and the surface of the body being orbited, i.e., the sun.

The By “mechanical” I am assuming you mean kinetic.

KE = 1/2 mv2

m = mass

v = the velocity of the orbiting body

E = 1/2 mv2

The velocity of the earth around the sun varies but the average is 29.783 kilometers per second.

The mass of the earth is 5.98 Ã— 10^24 kg. (the 24 is a power of the ten)

E = 0.5(5.98 Ã— 1024 kg)(29.783)2 = 3 x 10^24 x 900 = 2.7 x 10^27 joules