Bjarne said:
I guess it must be a mistake, because when gravity (suddenly) decrease, the Earth would fall closer to the Sun. (and hence a shorter period). Anyway I understand what you mean.
No, the orbit would increase to a longer period. Consider this:
A gravity decrease of 10% is the equivalent of a mass decrease of 10% in the mass of the Sun. With my example of the Earth, the present orbital velocity is 30 km/sec. The circular orbital velocity at the present Earth-Sun distance for a Sun 90% of its present mass would
be
V_o = \sqrt{\frac{GM}{a}}
where
a = 1.496e11 meters
M = 1.8e30 Kg (90% of solar mass)
G = 6.6733-11 m³/ kg/s²
This give an answer of 28.3 km/sec. Which is less than the Earth's velocity. This puts the Earth is the same situation as a body at the perihelion of an elliptical orbit.
To find the parameters of this orbit, we use energy conservation.
The energy of the Earth is the sum of its kinetic energy and gravitational potential or:
E = \frac{mv^2}{2}- \frac{GMm}{r}
v = 30,000 m/s
"m" being the mass of the Earth
"r" being the Earth-sun distance (1.496e11 m)
The energy can also be expressed as:
E = - \frac{GMm}{2a}
where "a" is the semi-major axis of the orbit, or average orbital distance.
Combining these two equations and solving for "a" gives us ~170 million km for the semi-major axis of the new orbit.
Taking the difference between the Earth's present distance and semi-major axis and adding this difference to the semi- major axis, gives us the new aphelion of the orbit.
Using
T = 2 \pi \sqrt{\frac{a^3}{GM}}
we find the period of the orbit.
Using the vis a vis equation:
v^2=\mu\left({{2 \over{r}} - {1 \over{a}}}\right)
we get the orbital velocity at aphelion.
By the way:
When radius (in a gravitation field) increase 100 times,
then (off course) acceleration du to gravity decreases 10.000 times (r^2) (100*100)
BUT the orbit velocity only decreases 10 times, - why not more than that?
I wonder: why gravity is 1000 times weaker than the orbit velocity.
How is the orbit velocity and force of gravity "connected?"
For a circular orbit, you can consider how much centripetal force is needed to maintain the circular path of the planet vs the gravitational force. ( do not put too much into this consideration as it is only useful when dealing with circular orbits.)
Thus centripetal force is found by
F_c = \frac{mv^2}{r}
Note that as r increases, the velocity needed to keep the centripetal force constant also decreases.
Gravitational force is found by
F_g = \frac{GMm}{r^2}
If we make Fc=Fg and solve for v we get:
v= \sqrt{\frac{GM}{r}}
Which is the equation I used above, where I used a for r.
Note that the orbital velocity decreases by the square-root of distance. All other things staying the same, a factor of 10 increase in orbit radius results in a factor of 3.16 decrease in orbital velocity, not a factor of 10 decrease.
