Where can a satellite orbit the sun at the same rate as the Earth does?

On Thu, 24 Jun 1999, Michael wrote:

> Hello! There's a satellite that is either in orbit or is going to be in
> orbit directly between the Earth and Sun, having the same angular
> velocity as the Earth. I calculate a distance between the Sun and
> Satellite of 1.469*10^11meters, or 2.1*10^9meters from the Earth. Is
> this the placement of this satellite?

Hello,

If the Earth had no gravitational field, then the satellite would have the
same angular velocity as the Earth at the distance of Earth's orbit from
the sun. If it were closer to the sun, it would have a higher angular
velocity (i.e. complete and orbit more quickly) and if it were further
away, it would have a lower angular velocity.

However, the Earth does have gravity (thankfully for us!), and therefore
there are a few points where a satellite can orbit with the same angular
velocity as the Earth. These are called the Lagrange points, of which
there are five. A schematic plot of these five (along with an intense
mathematical description of orbits around a Lagrange point) can be found
at the following web site:
http://www.treasure-troves.com/physics/LagrangePoints.html

The location of the lagrange point is where the combined gravitational
force of the sun and the Earth is equal to the centripetal force required
to keep a body in motion. For simplicity, let us assume that you are
interested in the L1 point, which falls on a line between the Earth and
the sun. The gravitational force due to the Earth is:
G Me Ms / d^2
where G is the gravitational constant, Me is the mass of the earth, Ms is
the mass of the satellite, and d is the distance from the earth.
The gravitational force due to the sun is
G Msun Ms / (Re - d)^2
where Msun is the mass of the sun and Re is the Earth-sun distance.

The centripetal force on the satellite is:
Ms v^2 / (Re - d)
where v is the orbital speed of the satellite around the sun. This
velocity is equal to the distance the satellite must go in one year
divided by the length of a year, or v = 2*pi*(Re - d) / t, where t is the
length of a year. filling in, the centripetal force becomes
4 pi^2 Ms (Re - d) / t^2
Now, set the gravitational forces (which are in opposite directions, so we
subtract them) equal to the centripetal force:
G Msun Ms G Me Ms 4 pi^2 Ms (Re-d)
--------- - ------- = ----------------
(Re-d)^2 d^2 t^2
Solve this for d, the distance the satellite is for the Earth, and,
assuming I did my math correctly, I get (after filling in the known
values)
d=1.55*10^11 cm
or 1.5 * 10^9 meters, which agrees with your calculation (accounting for
my rounding of masses). Good work!

Sincerely,
Kurtis Williams

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