Sect 6 4 Satellites and Weightlessness Satellites are

Sect. 6 -4 Satellites and “Weightlessness”

Satellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it “escapes” Earth’s gravity altogether. What keeps a satellite in orbit? The centripetal acceleration is CAUSED by the Gravitational Force! F = (mv 2)/r = G(m. ME)/r 2 The satellite’s speed or centripetal acceleration keeps it in orbit!! So, the satellite is kept in orbit by its speed—it is continually falling, but the Earth curves from underneath it.

The satellite is kept in orbit by its speed—it is continually falling, but the Earth curves from underneath it. Newton’s 1 st Law: If there were no (gravitational) force, the satellite would move in a straight line! “Falls” in a circle

Example 6 -6: Geosynchronous Satellite A geosynchronous satellite is one that stays above the same point on the Earth, which is possible only if it is above a point on the equator. Such satellites are used for TV and radio transmission, for weather forecasting, and as communication relays. Calculate: a. The height above the Earth’s surface such a satellite must orbit b. The satellite’s speed. c. Compare to the speed of a satellite orbiting 200 km above Earth’s surface. r

Conceptual Example 6 -7: Catching a satellite. You are an astronaut in the space shuttle pursuing a satellite in need of repair. You find yourself in a circular orbit of the same radius as the satellite, but 30 km behind it. How will you catch up with it?

“Weightlessness” • Objects in a satellite, including people, experience “EFFECTIVE weightlessness”. What causes this? • NOTE: THIS DOES NOT MEAN THAT THE GRAVITATIONAL FORCE ON THEM IS ZERO! They still have a gravitational force acting on them! • The satellite & all its contents are in “free fall”, so there is no normal force. This is what leads to the experience of “weightlessness”. More properly, this effect should be called apparent or effective weightlessness, because the gravitational force still exists. • To understand this, first, look at a simpler problem: A person in an elevator. (4 cases)

• Case 1: Person in elevator. Mass m attached to scale. No acceleration. (no motion) (a = 0) ∑F = ma = 0 • W = weight = upward force on mass m. By Newton’s 3 rd Law, equal & opposite to reading on scale. • ∑F = 0 = W - mg Scale reading is W = mg

• Case 2: Elevator moves up with acceleration a. ∑F = ma or W - mg = ma • W = upward force on m. Newton’s 3 rd Law equal & opposite to the scale reading. Scale reading (apparent weight) W = mg + ma > mg ! For a = (½)g, W = (3/2)mg Person is “heavier” also! Person experiences 1. 5 g’s!

• Case 3: Elevator moves down with acceleration a. ∑F = ma or W - mg = -ma • W = upward force on m. Newton’s 3 rd Law equal & opposite to scale reading. Scale reading (apparent weight) (½)mg W = mg - ma < mg ! For a = (½)g, W = (½)mg Person is “lighter” also! Person experiences 0. 5 g’s! (down)

• Case 4: Elevator cable breaks! Free falls down with acceleration a = g! ∑F = ma or W - mg = -ma, W = upward force on m. Newton’s 3 rd Law equal & opposite to scale reading. Scale reading (apparent weight) W = mg -ma = mg - mg W = 0! Elevator & person are apparently “weightless”. Clearly, though, THE FORCE OF GRAVITY STILL ACTS!

So, this effect of apparent weightlessness obviously doesn’t mean that gravity is not there, because the gravitational force obviously still exists. Other than in a falling elevator, it can be experienced in other ways on Earth, but only briefly:

• (Apparent) “weightlessness” in satellite orbit? • With gravity, satellite (mass ms) free “falls” continually! Just so gravitational force = centripetal force. F = G(ms. ME)/r 2 = msv 2/r • Consider scale in satellite: ∑F = ma = -mv 2/r (towards Earth center!) W - mg = -mv 2/r Or, W = mg -mv 2/r << mg! “Falls” in a circle

LESSON!!! • TV reporters are just plain WRONG (!!) when they say things like: “The space shuttle has escaped the Earth’s gravity”. Or “has reached a point outside the Earth’s gravitational pull”. • Why? Because the gravitational force is F = G(ms. ME)/r 2 This exists (& is NOT zero) even for HUGE distances r away from Earth! F 0 only for r