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Section 5-8: Satellites and “Weightlessness”

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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(mM E )/r 2 Its speed or it’s 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.

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

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Example 5-14: 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. r b. The satellite’s speed. c. Compare that to the speed of a satellite orbiting 200 km above Earth’s surface. Earth, mass M E m

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“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 F N. 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. This can be experienced on Earth also, but only briefly.

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Case 1 A person in an elevator. A mass m is attached to a scale. There is no acceleration. (no motion) (a = 0) ∑F = ma = 0 W = “weight” = upward force on mass m. By Newton’s 3 rd Law, W is equal & opposite to the reading on the scale. ∑F = 0 = W – mg Scale reading is W = mg To understand this, lets first, look at a simpler problem: A person riding an elevator (in 4 different cases).

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Case 2 The elevator moves up with acceleration a. ∑F = ma or W - mg = ma W = upward force on m. By Newton’s 3 rd Law, W is equal & opposite to the reading on the scale. 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!

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Case 3 Elevator moves down with acceleration a. ∑F = ma or W - mg = -ma W = upward force on m. By Newton’s 3 rd Law, W is equal & opposite to the reading on the scale. Scale reading (apparent weight) W = mg - ma < mg ! For a = (½)g, W = (½)mg Person is “lighter” also! Person experiences 0.5 g’s! (½)mg (down)

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

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

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

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