# Orbit velocity question

1. Sep 12, 2009

### Bjarne

Let say gravity suddenly decreased 10%, - for instance due to tidal friction etc...
How fast would a planet decelerate? -
How can this be calculated?

2. Sep 12, 2009

### Janus

Staff Emeritus
What effect would tidal friction have on gravity?

If you are asking what would happen to a planet's orbit if the gravity between it and the Sun suddenly(instantly) decreased by 10%, then it would enter a new orbit with its present distance at perihelion. For instance, the Earth would enter a orbit with a perihelion at 149.6 million km and an aphelion at 1.91 million km, with a period of 446.5 days.

In its climb to aphelion, it will slow to 23.5 km/sec from its present 30 km/sec, so in 223.25 days it will slow 6.5 km/sec. Off course it will regain that velocity as it falls back to perihelion.

3. Sep 12, 2009

### Bjarne

Thanks a lot.
Where did you got the (basis) data for the Earths, used above ?

4. Sep 12, 2009

### Bjarne

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.

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?"

Last edited: Sep 12, 2009
5. Sep 12, 2009

### Janus

Staff Emeritus
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.
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.

Last edited: Sep 12, 2009
6. Sep 12, 2009

### acr

Isn't it

$$E = - \frac{GMm}{2a}$$

?

The virial theorem states that

$$<E> = 1/2<U>$$

and it can be shown that

$$<U> = - \frac{GMm}{a}$$

7. Sep 12, 2009

### Janus

Staff Emeritus
Yes, a LaTex typo on my part. I will go back and fix it.

8. Sep 13, 2009

### Bjarne

Thanks a lot.
That was really useful.