Transcript Satellites & “Weightlessness”

Satellites & “Weightlessness”
“Weightlessness”
• What keeps a satellite in orbit?
The centripetal acceleration
is CAUSED by the
Gravitational Force!
F = (mv2)/r = G(mME)/r2
Its speed or centripetal
acceleration keeps it
in orbit!!
• Newton’s 1st Law: If no there were (gravitational)
force, the satellite would move in a straight line!
“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!
• 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 3rd
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 3rd 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 3rd 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 3rd 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!
• (Apparent) “weightlessness” in satellite orbit?
• With gravity, satellite
(mass ms) free “falls”
continually! Just so
gravitational force =
centripetal force.
F = G(msME)/r2 = msv2/r
• Consider scale in satellite:
∑F = ma = -mv2/r
(towards Earth center!)
W - mg = -mv2/r
Or, W = mg -mv2/r << mg!
• 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(msME)/r2
This exists (& is NOT zero) even for HUGE
distances r away from Earth!
F  0 only
for r  