Form 6 physics

From principle involved in circular motion, the satellite requires a centripetal force in order to maintain a stable orbit above the earth surface. This centripetal force is given by m.v^2/R, where m is the satellite mass, v is the satellite speed, and R is the distance of the orbit from the centre of the earth.

The required centripetal force is provided by earth's gravitational attraction, which is described by Newton's Law of Gravitation, GMm/R^2, where M is the earth mass and G is a constant (the Gravitational constant).

Hence, GMm/R^2 = mv^2/R
simplifying, we get v^2 = GM/R
or v = square-root[GM/R]

The speed of satellite in a stable orbit is thus inversely proportional to the sqaure-root of the distance of the orbit from earth centre.
The higher the altitude of the satellite, the larger the value of R, hence a lower speed v for the satellite.

2009-11-15 14:36:28 補充：
The physics principle behind is that at higher altitude, the earth's gravitational force becomes weaker, if the satellite speed is high, the earth's attractive force would not be strong enough to hold the satellite in orbit, it would flies away from the earth.

2009-11-15 14:36:38 補充：
Hence, the satellite's orbital speed must be low enough in order to tally with the decreased gravitational attraction.

By Newton's law of Gravitation,

Gravitational force between the satellite and the Earth, F

= GMm / R^2

G is Universal Gravitational constant

M is the mass of the Earth

m is the mass of the satellite

R is the distance between the satellite and the centre of the Earth

For this force, it accounts for the centripetal force required for the satellite to orbit around the Earth.

So, F = mv^2 / R

Therefore,

mv^2 / R = GMm / R^2

v = sqrt[GM / R]

We see that for higher altitude, that is the distance between the satellite and the centre of the Earth is larger (R ↑), then the value of v is smaller.

So, it has a smaller orbital speed.