Although g may be about 9.8 m/s^2 at the surface of earth, it is not 9.8 at 4 Re and therefore you can not assume it is constant
g = 9.8 (Re^2/r^2)
with what velocity must a body be thrown from earth surface so that it may reach a height 4Re the earth surface. given Re=6400km g=9.8m/s
2 answers
now
F = m g
work done against gravity = integral F*dr or m g dr
or
m * integral g dr
m 9.8 Re^2 integral dr/r^2
from Re to 4 Re
integral dx/x^2 = -1/x
so here
1/6,400,000 - 1/25,600,000
= 1.17 * 10^-7
so work done in Joules -
m 9.8 Re^2 * 1.17* 10^-7
= m (47,040,000) Joules
THAT = (1/2)mv^2
so
v = 9,700 meters/second
F = m g
work done against gravity = integral F*dr or m g dr
or
m * integral g dr
m 9.8 Re^2 integral dr/r^2
from Re to 4 Re
integral dx/x^2 = -1/x
so here
1/6,400,000 - 1/25,600,000
= 1.17 * 10^-7
so work done in Joules -
m 9.8 Re^2 * 1.17* 10^-7
= m (47,040,000) Joules
THAT = (1/2)mv^2
so
v = 9,700 meters/second
1 answer