Resource Lesson
Advanced Gravitational Forces
Printer Friendly Version
The
Law of Universal Gravitation
states that every object in the universe attracts every other object in the universe with a force that has a
magnitude
which is directly proportional to the product of their masses and inversely proportional to the distance between their centers squared.
where
G is the gravitational constant, 6.67 x 10
^{-11}
Nm
^{2}
/kg
^{2}
M
_{1}
is the mass of the first body in kg
M
_{2}
is the mass of the second body in kg
R is the distance from the center of M
_{1}
to the center of M
_{2}
In this lesson we will look at some special cases of universal gravitation whose calculations require the use of more sophisticated math and sometimes, integral calculus.
Force between a point mass and an extended mass
Our first example involves calculating the gravitational force between a point mass
M
and an extended rod of mass
m
, length L, and mass per unit length,
.
To begin, divide the rod into a finite (countable) number of segments of mass
each located at a distance
x
from
M
.
Each of these segments will contribute a gravitational force of attraction.
If we take the limit as
approaches zero, then our expression for
F
becomes
Before we can integrate we must express
in terms of
.
Substituting and integrating gives us
where
.
Force between a point mass and two equidistant particles
Our next investigation will be to calculate the gravitational force exerted by two equal, equidistant massive particles,
M
on a point mass,
m
, located at point P.
In setting up this diagram, we place the x- and y-axes to reflect the problem's symmetry.
Since the y-components of these attractive forces will cancel,
we will sum up the two x-components to determine our answer.
Using the formula for universal gravitation and the distances labeled in the above diagram
we can calculate our equation for net F
_{x}
.
Notice that this net gravitational force on our point mass,
m
,
points to the origin, or the center of the
2M
system.
Our result will allow us to calculate the gravitation force at any position P along the x-axis. In particular, notice that
when x = 0, the net force will become zero; that is, the point mass,
m
, would feel no gravitational force.
Force on a point mass located along the axis of a thin ring
Suppose we now allow the two masses in the previous example to be rotated about the x-axis to create a thin ring in the yz-plane having a radius of
a
and a mass of
2M
. To accomplish this let each mass "create" half of the ring. Our point mass,
m
, is now located on the axis of our thin ring. How will this continuous distribution of mass affect our previous result?
Now, imagine that the ring, like the bar in example #1 is made up of little segments
. If we pair up segments based on the requirement that they are on opposite sides of any selected diameter, then we can calculate the force the ring exerts on the point mass,
m
.
Each pair of
segments will be the same distance
r
from
m
. No matter where the mass segment is located on the ring, only the force components pointing along the x-axis, towards the center of the ring at the origin, will contribute towards the force of gravitational attraction between the ring and the point mass. This symmetry will allow us to calculate the gravitational force on the mass by the ring.
As before, using the formula for universal gravitation and the distances diagram shown above, note that:
Adding up both x-components that point to the center of the ring will yield the following expression for
net F
_{center}
. Note that we must also pass to the limit to convert our summation into the desired integral.
Before we can integrate we must express
dM
in terms of the arc length,
s
, of the ring. This arc length for the entire ring will equal the ring's circumference.
This substitution along with our original supposition that "each mass,
M
" was stretched out to form
half of our ring
will allow us to set up the limits of the integral and evaluate it.
As we complete our work also notice that
G
,
m
,
a
,
x
, and
are all constants and can be removed from the integral.
Our last step is to find an expression for
M
and replace it.
Notice that this is the EXACT same expression that we calculated for the net gravitational force in the previous example involving two equidistant particles. Stretching the masses,
M
, out into a ring did NOT affect the magnitude of the gravitational force on the point mass.
Final Steps
At this junction in the lesson our goal to determine what the gravitational force would be on a point mass,
m
, if the point mass were to be placed (1)
outside of a thin spherical shell
, (2)
inside a thin spherical shell
, and then (3)
inside a solid spherical mass
. Our previous examples have paved the path for these investigations.
(1) First we will analyze the force that would be exerted on a point mass,
m
, if it were to placed outside of a thin spherical shell at a distance
x
from the shell's center. To do this, we can consider our uniform, spherical shell to be made up of thin rings symmetric to the x-axis. Each ring contributing to the shell's final total mass,
M
. The rings are going to increase in radius, from r = 0 to r = a, and then decrease back to 0 as we take our slices.
Complications arise from the number of variables that need to be related and then integrated. As you can see, the rings will have different radii and circumferences, have different values for angles θ and
α
, as well as different values of r and x. For those of you interested in this advanced solution, a derivation can be found in section 11-5 of your text (Tipler,
Physics for Scientists and Engineers
, 5th ed, pages 358-360).
Although the actual calculus for this derivation is beyond the scope of our lesson, remember that in general for a thin ring the gravitational attraction between it and an external point mass,
m
, is directed along the x-axis towards the center of the ring.
The conceptual understanding that we are adding together rings with forces all pointing towards the center of the shell will help us rationalize that the external point mass,
m
, "would see a massive shell" as if it were a concentrated point mass,
M
, located at the center of the shell and exerting a force
directed towards it center. This derivation can be revisited once we understand Gauss' Law and use its gravitational analogue to prove our result more simply.
If our point mass,
m
, were a distance
x
from a solid sphere, we would expect the same outcome since we can envision our solid sphere as being composed of tightly nested shells. What makes this result all the more
beautiful
is that it shows that the attractive force between the Earth and a softball arching through the air does not need to become a complicated calculation depending on local terrain. The gravitational force is between the mass of the softball,
m
, and the mass of the Earth,
M
_{E}
, as if it were a point mass at concentrated at the Earth's center. A truly amazing, simple result!
(2) To calculate the gravitational force on a point mass,
m
, placed inside a thin spherical shell at any point P, we will use properties of similar right circular cones along with Newton's Law of Universal Gravitation.
Through point P, draw two intersecting chords that terminate on opposite sides of our thin shell, crossing at the center of
m
. Our chords, with their equal apex angles, have created two similar right circular cones having base areas A
_{1}
and A
_{2}
. The mass segments represented by these areas are gravitationally attracted to our point mass,
m
. But what is the relationship between these areas?
Using the trigonometry ratio tangent, we can develop a relationship between the heights of each cone and the radii of their circular bases.
Gravitationally, these base areas each exert a force of attraction on our point mass,
m
.
where m
_{1}
represents the mass of area A
_{1}
, and m
_{2}
represents the mass of area A
_{2}
. If we let σ represent a uniform mass per unit area, kg/m
^{2}
, for a cross-sectional area of our shell, then m
_{1}
= σA
_{1}
and m
_{2}
= σA
_{2}
.
As you can see, the forces on
m
are exactly equal and would cancel. If we repeated this process for every surface segment, we would always discover that the forces are always balanced. That is, our point mass would feel NO gravitational force at point P, no matter where P is inside the shell.
(3) To calculate the gravitational force on a point mass,
m
, located at a radius
r
inside a solid uniform sphere of radius
R
, we will need to use our outcome for the gravitational force experienced inside a thin shell as well as our knowledge of density.
In the diagram given above, the point mass,
m
, is represented by the white dot. All of the sphere's mass above its location can be considered to consist of nested shells (shown in gray). From our earlier discussions, we know that the gravitational force on a point mass located anywhere inside a shell equals zero. Therefore we can discount their masses and only examine the mass for radii 0 <= r.
Using the fact that the sphere has a uniform density, ρ, in kg/m
^{3}
, we can express the mass of the interior sphere in terms of the total mass of the entire sphere.
This leads us to our final substitution showing that the gravitational force between the interior mass and our point mass is proportional to r's percentage of the original radius R.
A graph summarizing the results for a solid sphere is given below.
Related Documents
Lab:
Labs -
Coefficient of Friction
Labs -
Coefficient of Friction
Labs -
Coefficient of Kinetic Friction (pulley, incline, block)
Labs -
Conservation of Momentum in Two-Dimensions
Labs -
Falling Coffee Filters
Labs -
Force Table - Force Vectors in Equilibrium
Labs -
Gravitational Field Strength
Labs -
Inelastic Collision - Velocity of a Softball
Labs -
Inertial Mass
Labs -
Kepler's 1st and 2nd Laws
Labs -
Lab: Triangle Measurements
Labs -
LabPro: Newton's 2nd Law
Labs -
Loop-the-Loop
Labs -
Mars' Lab
Labs -
Mass of a Rolling Cart
Labs -
Moment of Inertia of a Bicycle Wheel
Labs -
Relationship Between Tension in a String and Wave Speed
Labs -
Relationship Between Tension in a String and Wave Speed Along the String
Labs -
Static Equilibrium Lab
Labs -
Static Springs: Hooke's Law
Labs -
Static Springs: Hooke's Law
Labs -
Static Springs: LabPro Data for Hooke's Law
Labs -
Terminal Velocity
Labs -
Video LAB: A Gravitron
Labs -
Video LAB: Ball Re-Bounding From a Wall
Labs -
Video Lab: Falling Coffee Filters
Resource Lesson:
RL -
Advanced Satellites
RL -
Air Resistance
RL -
Air Resistance: Terminal Velocity
RL -
Forces Acting at an Angle
RL -
Freebody Diagrams
RL -
Gravitational Energy Wells
RL -
Gravitational Potential Energy
RL -
Inclined Planes
RL -
Inertial vs Gravitational Mass
RL -
Newton's Laws of Motion
RL -
Non-constant Resistance Forces
RL -
Properties of Friction
RL -
Springs and Blocks
RL -
Springs: Hooke's Law
RL -
Static Equilibrium
RL -
Systems of Bodies
RL -
Tension Cases: Four Special Situations
RL -
The Law of Universal Gravitation
RL -
Universal Gravitation and Satellites
RL -
Universal Gravitation and Weight
RL -
What is Mass?
RL -
Work and Energy
Review:
REV -
Review: Circular Motion and Universal Gravitation
Worksheet:
APP -
Big Fist
APP -
Family Reunion
APP -
The Antelope
APP -
The Box Seat
APP -
The Jogger
CP -
Action-Reaction #1
CP -
Action-Reaction #2
CP -
Equilibrium on an Inclined Plane
CP -
Falling and Air Resistance
CP -
Force and Acceleration
CP -
Force and Weight
CP -
Force Vectors and the Parallelogram Rule
CP -
Freebody Diagrams
CP -
Gravitational Interactions
CP -
Incline Places: Force Vector Resultants
CP -
Incline Planes - Force Vector Components
CP -
Inertia
CP -
Mobiles: Rotational Equilibrium
CP -
Net Force
CP -
Newton's Law of Motion: Friction
CP -
Static Equilibrium
CP -
Tensions and Equilibrium
NT -
Acceleration
NT -
Air Resistance #1
NT -
An Apple on a Table
NT -
Apex #1
NT -
Apex #2
NT -
Falling Rock
NT -
Falling Spheres
NT -
Friction
NT -
Frictionless Pulley
NT -
Gravitation #1
NT -
Head-on Collisions #1
NT -
Head-on Collisions #2
NT -
Ice Boat
NT -
Rotating Disk
NT -
Sailboats #1
NT -
Sailboats #2
NT -
Scale Reading
NT -
Settling
NT -
Skidding Distances
NT -
Spiral Tube
NT -
Tensile Strength
NT -
Terminal Velocity
NT -
Tug of War #1
NT -
Tug of War #2
NT -
Two-block Systems
WS -
Advanced Properties of Freely Falling Bodies #1
WS -
Advanced Properties of Freely Falling Bodies #2
WS -
Advanced Properties of Freely Falling Bodies #3
WS -
Calculating Force Components
WS -
Charged Projectiles in Uniform Electric Fields
WS -
Combining Kinematics and Dynamics
WS -
Distinguishing 2nd and 3rd Law Forces
WS -
Force vs Displacement Graphs
WS -
Freebody Diagrams #1
WS -
Freebody Diagrams #2
WS -
Freebody Diagrams #3
WS -
Freebody Diagrams #4
WS -
Introduction to Springs
WS -
Kepler's Laws: Worksheet #1
WS -
Kepler's Laws: Worksheet #2
WS -
Kinematics Along With Work/Energy
WS -
Lab Discussion: Gravitational Field Strength and the Acceleration Due to Gravity
WS -
Lab Discussion: Inertial and Gravitational Mass
WS -
net F = ma
WS -
Parallel Reading - The Atom
WS -
Practice: Vertical Circular Motion
WS -
Ropes and Pulleys in Static Equilibrium
WS -
Standard Model: Particles and Forces
WS -
Static Springs: The Basics
WS -
Universal Gravitation and Satellites
WS -
Vocabulary for Newton's Laws
WS -
Work and Energy Practice: Forces at Angles
TB -
Systems of Bodies (including pulleys)
TB -
Work, Power, Kinetic Energy
PhysicsLAB
Copyright © 1997-2018
Catharine H. Colwell
All rights reserved.
Application Programmer
Mark Acton