3 ms·
hmmm. mass of earth is M=6e24 kg, radius R=6.4e6 m, G=6.7e-11 m^3 kg^-1 s^-2. so the "energy debt" on the surface (the gpe) is GM/R=6.3e7 J. he's equating th
by andrewcooke 17y ago
hmmm. mass of earth is M=6e24 kg, radius R=6.4e6 m, G=6.7e-11 m^3 kg^-1 s^-2.
so the "energy debt" on the surface (the gpe) is GM/R=6.3e7 J.
he's equating that to an equivalent height H for 1kg in a constant field of 9.8ms^-2, so H=6.4e6 m. so the 6,400 km number is correct (the inset) (curiously, this number is also the radius of the earth, approx).
perhaps the value in the main figure is a mistake, or he is folding in some other effect (i can't work out what, though).
edit: i see that he gives a formula in the well for jupiter, which is equivalent to what i just did, so i think the main diagram's value for earth is a mistake.
- cperciva 17y agoso the 6,400 km number is correct (the inset) (curiously, this number is also the radius of the earth, approx). It's not curious, and it's not approximate. The gravitational potential at distance r is - G M_1 M_2 / r, and the gravitational force at distance r is G M1 M2 / r^2; so you will always need to overcome exactly one radius of surface gravity.
- deleted 17y ago[deleted]
- andrewcooke 17y agooh, yeah, right.