2Its speed or it’s centripetal acceleration keeps it in orbit!! 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? Thecentripetal acceleration is CAUSED by the Gravitational Force!F = (mv2)/r = G(mME)/r2Its speed or it’s centripetal acceleration keeps it in orbit!!So, the satellite is kept in orbit byits speed—it is continually “falling”, but the Earth curves from underneath it.Figure Caption: A satellite, the International Space Station, circling the Earth.Figure Caption: Artificial satellites launched at different speeds.
3The satellite is kept in orbit by its speed it is continually falling, but the Earth curves from underneath it.Newton’s 1st Lawtells us that if there wereno (gravitational) force,the satellite would movein a straight line!“Falls” ina circle!Figure Caption: A moving satellite “falls” out of a straight-line path toward the Earth.
4Example 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’ssurface such a satellite must orbit.mrEarth, mass MEb. The satellite’s speed.c. Compare that to the speed of a satellite orbiting 200km above Earth’s surface.
5“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 FN. 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.
6the reading on the scale. To understand this, lets first, look at a simpler problem:A person riding an elevator (in 4 different cases).Case 1A person in an elevator.A mass m is attached to a scale.There is no acceleration. (no motion)(a = 0) ∑F = ma = 0W = “weight” = upward forceon mass m. ByNewton’s 3rd Law,W is equal & opposite tothe reading on the scale.∑F = 0 = W – mg Scale reading is W = mg
7the reading on the scale. Case 2The elevator moves upwith acceleration a. ∑F = ma or W - mg = maW = upward force on m. ByNewton’s 3rd Law,W is equal & opposite tothe reading on the scale. Scale reading(apparent weight)W = mg + ma > mg !For a = (½)g, W = (3/2)mgPerson is “heavier” also!Person experiences 1.5 g’s!
8the reading on the scale. Case 3Elevator moves down with acceleration a. ∑F = ma or W - mg = -maW = upward force on m. ByNewton’s 3rd Law,W is equal & opposite tothe reading on the scale. Scale reading(apparent weight)W = mg - ma < mg !For a = (½)g, W = (½)mgPerson is “lighter” also!Person experiences 0.5 g’s!(½)mg(down)
9the reading on the scale. Case 4The elevator cable breaks! It free falls down with acceleration a = g! ∑F = ma or W - mg = -ma,W = upward force on m. ByNewton’s 3rd Law,W is equal & opposite tothe reading on the scale. Scale reading(apparent weight)The elevator & person are apparently “weightless”. Clearly, though,THE FORCE OFGRAVITY STILL ACTS!
10F = G(msME)/r2 = (msv2)/r ∑F = ma = - (mv2)/r W - mg = - (mv2)/r (Apparent) “weightlessness” in the satellite orbit?With gravity, the satellite(mass ms) free “falls”continually! Just so thegravitational force = thecentripetal force.F = G(msME)/r2 = (msv2)/rConsider a scale in the satellite:∑F = ma = - (mv2)/r(towards the Earth’s center)W - mg = - (mv2)/rOr, W = mg - (mv2)/r << mg!“Falls” ina circle!
11LESSON. TV reporters are just plain WRONG ( 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 isF = G(msME)/r2This exists (& is NOT zero) even for HUGE distances r away from Earth!F 0 ONLY for r