Download Chapter 13: Universal Gravitation

Chapter 13: Universal Gravitation
Hannah Humayun
Summary of Key Points
-Newton’s Law of Universal
Gravitation: “Every particle in the
universe attracts every other particle
with a force proportional to the
product of their masses, and inversely
proportional to the square of the
distance between them.”
-If the system is bound then the energy
is negative
-Potential Energy decreases as an
object get closer to the Earth’s surface
- Fc = Fg
-Acceleration is the measure of the
gravity field
-Kepler’s Law: All planets move in
elliptical orbits with the Sun as a foci
Equations !! !!
! !
Fg = -­‐G ! ! ř or Fg = -­‐G !! ! ! For the force of gravity
between two objects
!!
Fc = m !
Centripetal Acceleration
W = ∫ 𝐹! 𝑑𝑥
Work done by the force of
gravity
Important Numbers:
!
-∆ 𝑈 = ∫! ! 𝑚𝑔 𝑑𝑥
!
Work is equal to the negative
change in potential energy
Etotal = KE + U
Total energy with regards to
Kinetic and Potential Energy
Helpful Vocabulary Terms
Perihelion: The point of an orbit that is closest to the sun
Aphelion: The point in an orbit that is farthest from the sun
Weight: Force exerted on an object by gravity (mg)
Mass: quantity of matter in an object
REarth = 6.38 x 106 m
Rmoon’s orbit = 3.85 x 108 m
Mass of Earth: 5.98 x 1024 kg
G = 6.672 x 10-11
!! !
!" !
Moon Orbit time: 27.3 days
The Scientists Behind Universal Gravitation Kepler (1561-­‐1630): Started out as an assistant to Brahe but then moved on to analyze the motion of the planets mathematically. Satellite: a celestial body orbiting the earth or another planet
Geosynchronous: period of rotation the same as Earth’s
Galileo (1564-­‐1642): Observed the universe through a telescope. Newton (1642-­‐1727): Discovered that all objects attract one another and that force is related to the mass of the objects involved. Created the Law of Universal Gravitation. In this image, the highlighted red areas are the same by Kepler!
Now let’s try some problems regarding universal gravitation! J
Easy: Consider a satellite of mass 300 kg moving in a circular orbit
around the Earth at a constant speed v and at an altitude 1000 m
above the Earth’s surface. Determine the speed of the satellite (hint:
using G, height above Earth’s surface, radius of the earth, and mass
of the Earth).
Medium: If a ball is released from 20,000 m above the Earth’s surface,
how fast is it traveling just before it hits the ground? (For this problem, ignore air resistance factors)
Hard: Three 0.3 kg billiard balls are placed on a table at the corners of a right triangle as shown in the figure
below. The sides of the triangles are of lengths a=.40 m, b = 0.3 m, and c= 0.5 m. Calculate the gravitational
force vector on the cue ball (designated m1) resulting from the other two balls as well as the magnitude and
direction of this force.
Solutions:
Easy:
-The only external force acting on the satellite is the force of Earth’s gravity (which keeps the satellite in a
circular orbit). Therefore…
Fg = ma à -­‐G v=
!!!"#$!
!!"#$! !!
!! !!
!!
!!
= m !
à v =
This can be applied because of the centripetal nature of the orbit and the
force of gravity on the satellite. Solve for v with these equations.
!.!"# ! !"!!! (5.98 x 1024 )
!.!" ! !"!
907 m/s à
!(!""" !)
Medium:
To begin a problem like this, we will start with the conservation of energy equation.
𝐸!"#!$% = 𝐸!"#$%
Uintial + KEintial = Ufinal + KEfinal
Then continue to substitute for known values…
-
!"#
!
+0=-
!"#
!
+ ½ mv2
Move the potential energies to the same side of the equation and proceed to isolate velocity.
GMm ( !
v=
!
!"#!$%
− !
2𝐺𝑀 ( !
!
!"#$%
!
!"#!$%
) = ½ mv2
− !
!
!"#$%
)
Plug in values
v=
!
!
2 6.672 𝑥 10!!! (5.98 𝑥 10!" )( !.!" ! !"! − !.!" ! !"! !!",!!! )
625 m/s Hard:
This question is odd because normally we do not consider the
gravitational effects of everyday objects on one another because those
forces are typically very small. In addition, as this problem is dealing
with vector components, there are a lot of moving pieces in addition to
the forces being very small.
To start, we must find the force exerted by m2 on m1:
F=G
!! !!
!!
ĵ
= (6.672 x 10-11)
!.! !" (!.! !")
!.! !
ĵ
3.75 x 10-­‐11 ĵ N Then we find the force exerted by m3 on m1:
F=G
!! !!
!!
î
= (6.672 x 10-11)
!.! !" (!.! !")
!.!!
î
6.672 x 10-­‐11 î N Find the net gravitational force on m1 by adding the components you found above:
F = F21 + F31
= 3.75 x 10-­‐11 ĵ N + 6.672 x 10-­‐11 î N (6.672î, 3.75 ĵ) x 10-11 N
Proceed to find the magnitude of this force: (6.672)! + (3.75)! x 10-­‐11 7.66 x 10-­‐11 N For completion, find the tangent angle for the net force vector and evaluate: !
!.!" ! !"!!!
Tan (θ) = !!" = !.!"# ! !"!!! = 0.562
!"
θ = tan-1 (0.562) = 30°