Physics 1 Notes: Gravity, Work, Energy, Momentum

Chapter 6
• Newton’s Law of Universal Gravitation
o Gravity generated by any massive object
§ Is a vector force
§ Gravity is towards center of the massive object
o Force is attractive
o Force depends directly on the product of the masses
o Force is inversely proportional to the distance between the distance between
masses2




o
§ r = distance between centers
§ 𝐹⃗!" = −𝐹⃗"!
§ Gravitational force on m1 due to m2 is equal and opposite to force on m2
due to m1
#$ $
o 𝐹 = %!" "
§ G is gravitational constant
§ G=6.67*10-11 Nm2/kg2
o Ex: How much force is present between two socially-distanced people? Person 1
is 50kg, person 2 is 70kg, and they are 2m apart.
#$ $
§ 𝐹 = %!" "
('.')∗!+#!! )(-+)()+)
§ 𝐹= (")"
-8
§ 𝟓.8 * 10 N
§ Small because people are not massive
o Ex: How much force is present between the same person 1 and the Earth? The
mass of Earth is 5.98 * 1024 kg, and the distance from the surface of the earth to
the center is 6.38 * 106 m.
#$ $
§ 𝐹 = %!" "
('.')∗!+#!! )(-+)(-../∗!+"$ )
§ 𝐹= ('.0/∗!+% )"
§ 490 N
§ Can use this to find a due to gravity (g)
• Take force and divide it by mass
1
• 𝑎=$
2.+
• 𝑎= -+
• 9.8 m/s2

, #$3
• 𝐹= %"
= 𝑚𝑔
#3
• 𝑔 = %"
o g varies with r
§ Changes on Earth depending on altitude
#$&'()*
§ 𝑔45%67 = (% 8596:6;<4)"
&'()*
• Gravitational pull higher at higher altitude
• 1/distance2
§ Center of the Earth = 0g
§ Ex: Find g of the moon. The mass of the moon is 7.35 * 1022 kg.
('.')∗!+#!! )().0-∗!+"" )
• 𝑔$==> = (!.)2∗!+% )"
• 𝟏.62 m/s 2

o About 1/6 of gearth
o Satellites
§ Orbits something else
§ Moon is satellite of Earth




§
• Orbits achievable if tangential velocity is high enough
• Instead of falling down onto the Earth, it falls around it
• Acceleration becomes centripetal and centripetal force is gravity
#$3&'()* $? "
o 𝐹% = %"
= %
o Velocity needed to maintain orbital radius
#3
§ 𝑣 " = &'()* %
§ Ex: The ISS orbits at an altitude of 350km above the surface of the earth.
What velocity must it travel at to stay in orbit? How long does it take to
orbit?
• Velocity
o 𝑟 = 𝑟45%67 + 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒
o 𝑟 = 6.38 ∗ 10' 𝑚

, ('.')∗!+#!! )(-../∗!+"$ )
o 𝑣=: '.)0∗!+%
o 7700 m/s
• Time
<
o 𝑠̅ = 6
"@%
o 𝑠̅ = 6
"@%
o 𝑡= ?
"@('.)0∗!+% )
o 𝑡= ))++
o 5198 sec
o Escape velocity
§ What velocity is required to escape the orbit of a massive object
§ Has to do with energy, not force
§ 𝑣4AB = 𝑣√2
o Planets orbit around the sun
§ The sun has a gravitational pull




§
"@%
§ Orbital velocity: 𝑣 = 6
§ Derive the constant for objects orbiting the sun
• Substitute for velocity (M is mass of sun)
"@% " #3
• @ 6
A = %
2@" % " #3
• 6"
= %
6" 2@"
• %+
= #3
o Constant for objects orbiting the sun
o Newton’s Law of Universal Gravity
o Orbital period squared divided by orbital radius cubed is
constant for all objects orbiting the sun
5+
o 6"
= constant
§ a is avg. distance from the sun
§ t is period of orbit

, § Ex: How far is Mars from the sun compared to Earth? Mars period=1.88
years; Earth period=1 year; Earth distance from sun=150,000,000km
5+
• 6"
5, + 5- +
• 6, " = 6- "
5, +
• 𝑎3 0 = 6, "
∗ 𝑡3 "
! 5;+
• 𝑎3 0 = !" ∗ (1.88)"
• 𝑎3 0 = 1.52 𝑎𝑢
• 1.52 𝑎𝑢 = 1.52 ∗ 150000000𝑘𝑚
• 𝑎3 = 228000000𝑘𝑚

Chapter 7
• Work
o Displacement of an object times force exerted on the object in the direction of
displacement
§ MUST be same direction
o Reflects on acceleration
o 𝑊 = 𝐹𝑑
§ 𝑊 = 𝐹∆𝑥
o Conceptual
§ Start with an object of mass m at rest on a frictionless surface




§
• Net force is 0
§ Exert a force to move, it will accelerate




§
• Object moves through a displacement, d = ∆x, because of the
force