#$&*
course PHY 121
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
028. `query 28
PRELIMINARY STUDENT QUESTION:
Again I am still having trouble with what equations to use...for PE change we would use the equation PE = Gmm/r1 -
Gmm/r2 same as the Force equation or do we just find dPE by grav. Force * ds both are listed in my notes. As far as
KE I assume m*9.8m/s/s *rE/r^2
INSTRUCTOR RESPONSE
PE = -G M m / r^2 so going from radius r1 to radius r2 we would have `dPE = GMm/r1 - GMm/r2.
If r1 and r2 don't differ by much, then G M m / r1 and G M m / r2 won't differ by much, and it would be important to use an appropriate number of significant figures in calculating the difference.
Alternatively the difference can be expressed in terms of the common denominator r1 * r2 as G M m * ( r2 - r1) / (r2 * r1), where the common factor G M m has been factored out.
As another alternative if r1 and r2 are proportionally close, you can figure out the average force and multiply by `dr = r2 - r1, as described below:
`dPE, i.e., the change in gravitational PE, is also equal to average gravitational force * change in distance from planet center:
• `dPE = F_ave * `dr
This is valid if `dr is small compared to r1 and r2, for the following reasons and with the associated cautions:
Gravitational force is not linearly related to r (i.e., the graph of F vs. r is not a straight line), so the average force on an interval is not equal to the average of the initial and final force.
However if the proportional change in r is small enough the curvature of the graph won't be noticeable, and the average force could be approximated by the average of initial and final force. In fact it is possible that the proportional change in r is small enough that over the relevant interval there is no significant change in the force itself.
• In these cases the change in PE is more easily found using F_ave * `dr.
*********************************************
Question: `qQuery class notes #26
Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
`a** The proportionality is accel = k / r^2. When r = rE, accel = 9.8 m/s^2 so
9.8 m/s^2 = k / rE^2.
Thus k = 9.8 m/s^2 * rE^2, and the proportionality can now be written
accel = [ 9.8 m/s^2 * (rE)^2 ] / r^2. Rearranging this gives us
accel = 9.8 m/s^2 ( rE / r ) ^2, which we symbolize using g = 9.8 m/s^2 as
a = g rE^2 / r^2.
If we set the acceleration equal to v^2 / r, we obtain
v^2 / r = g ( rE / r)^2 so that
v^2 = g ( rE^2 / r) and
v = sqrt( g rE^2 / r) = rE sqrt( g / r)
Thus if we know the radius of the Earth and the acceleration of gravity at the surface we can calculate orbital velocities without knowing the universal gravitational constant G or the mass of the Earth.
If we do know G and the mass of the Earth, we can proceed as follows:
The gravitational force on mass m at distance r from the center of the Earth is
F = G m M / r^2,
Where M is the mass of the Earth and m the mass of the satellite. Setting this equal to the centripetal force m v^2 / r on the satellite we have
m v^2 / r = G m M / r^2, which we solve for v to get
• v = sqrt( G M / r).
**
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I just cannot seem to put it all together.
------------------------------------------------
Self-critique Rating:
#$&*
course PHY 121
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
028. `query 28
PRELIMINARY STUDENT QUESTION:
Again I am still having trouble with what equations to use...for PE change we would use the equation PE = Gmm/r1 -
Gmm/r2 same as the Force equation or do we just find dPE by grav. Force * ds both are listed in my notes. As far as
KE I assume m*9.8m/s/s *rE/r^2
INSTRUCTOR RESPONSE
PE = -G M m / r^2 so going from radius r1 to radius r2 we would have `dPE = GMm/r1 - GMm/r2.
If r1 and r2 don't differ by much, then G M m / r1 and G M m / r2 won't differ by much, and it would be important to use an appropriate number of significant figures in calculating the difference.
Alternatively the difference can be expressed in terms of the common denominator r1 * r2 as G M m * ( r2 - r1) / (r2 * r1), where the common factor G M m has been factored out.
As another alternative if r1 and r2 are proportionally close, you can figure out the average force and multiply by `dr = r2 - r1, as described below:
`dPE, i.e., the change in gravitational PE, is also equal to average gravitational force * change in distance from planet center:
• `dPE = F_ave * `dr
This is valid if `dr is small compared to r1 and r2, for the following reasons and with the associated cautions:
Gravitational force is not linearly related to r (i.e., the graph of F vs. r is not a straight line), so the average force on an interval is not equal to the average of the initial and final force.
However if the proportional change in r is small enough the curvature of the graph won't be noticeable, and the average force could be approximated by the average of initial and final force. In fact it is possible that the proportional change in r is small enough that over the relevant interval there is no significant change in the force itself.
• In these cases the change in PE is more easily found using F_ave * `dr.
*********************************************
Question: `qQuery class notes #26
Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
`a** The proportionality is accel = k / r^2. When r = rE, accel = 9.8 m/s^2 so
9.8 m/s^2 = k / rE^2.
Thus k = 9.8 m/s^2 * rE^2, and the proportionality can now be written
accel = [ 9.8 m/s^2 * (rE)^2 ] / r^2. Rearranging this gives us
accel = 9.8 m/s^2 ( rE / r ) ^2, which we symbolize using g = 9.8 m/s^2 as
a = g rE^2 / r^2.
If we set the acceleration equal to v^2 / r, we obtain
v^2 / r = g ( rE / r)^2 so that
v^2 = g ( rE^2 / r) and
v = sqrt( g rE^2 / r) = rE sqrt( g / r)
Thus if we know the radius of the Earth and the acceleration of gravity at the surface we can calculate orbital velocities without knowing the universal gravitational constant G or the mass of the Earth.
If we do know G and the mass of the Earth, we can proceed as follows:
The gravitational force on mass m at distance r from the center of the Earth is
F = G m M / r^2,
Where M is the mass of the Earth and m the mass of the satellite. Setting this equal to the centripetal force m v^2 / r on the satellite we have
m v^2 / r = G m M / r^2, which we solve for v to get
• v = sqrt( G M / r).
**
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I just cannot seem to put it all together.
------------------------------------------------
Self-critique Rating:
#*&!
#$&*
course PHY 121
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
028. `query 28
PRELIMINARY STUDENT QUESTION:
Again I am still having trouble with what equations to use...for PE change we would use the equation PE = Gmm/r1 -
Gmm/r2 same as the Force equation or do we just find dPE by grav. Force * ds both are listed in my notes. As far as
KE I assume m*9.8m/s/s *rE/r^2
INSTRUCTOR RESPONSE
PE = -G M m / r^2 so going from radius r1 to radius r2 we would have `dPE = GMm/r1 - GMm/r2.
If r1 and r2 don't differ by much, then G M m / r1 and G M m / r2 won't differ by much, and it would be important to use an appropriate number of significant figures in calculating the difference.
Alternatively the difference can be expressed in terms of the common denominator r1 * r2 as G M m * ( r2 - r1) / (r2 * r1), where the common factor G M m has been factored out.
As another alternative if r1 and r2 are proportionally close, you can figure out the average force and multiply by `dr = r2 - r1, as described below:
`dPE, i.e., the change in gravitational PE, is also equal to average gravitational force * change in distance from planet center:
• `dPE = F_ave * `dr
This is valid if `dr is small compared to r1 and r2, for the following reasons and with the associated cautions:
Gravitational force is not linearly related to r (i.e., the graph of F vs. r is not a straight line), so the average force on an interval is not equal to the average of the initial and final force.
However if the proportional change in r is small enough the curvature of the graph won't be noticeable, and the average force could be approximated by the average of initial and final force. In fact it is possible that the proportional change in r is small enough that over the relevant interval there is no significant change in the force itself.
• In these cases the change in PE is more easily found using F_ave * `dr.
*********************************************
Question: `qQuery class notes #26
Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
`a** The proportionality is accel = k / r^2. When r = rE, accel = 9.8 m/s^2 so
9.8 m/s^2 = k / rE^2.
Thus k = 9.8 m/s^2 * rE^2, and the proportionality can now be written
accel = [ 9.8 m/s^2 * (rE)^2 ] / r^2. Rearranging this gives us
accel = 9.8 m/s^2 ( rE / r ) ^2, which we symbolize using g = 9.8 m/s^2 as
a = g rE^2 / r^2.
If we set the acceleration equal to v^2 / r, we obtain
v^2 / r = g ( rE / r)^2 so that
v^2 = g ( rE^2 / r) and
v = sqrt( g rE^2 / r) = rE sqrt( g / r)
Thus if we know the radius of the Earth and the acceleration of gravity at the surface we can calculate orbital velocities without knowing the universal gravitational constant G or the mass of the Earth.
If we do know G and the mass of the Earth, we can proceed as follows:
The gravitational force on mass m at distance r from the center of the Earth is
F = G m M / r^2,
Where M is the mass of the Earth and m the mass of the satellite. Setting this equal to the centripetal force m v^2 / r on the satellite we have
m v^2 / r = G m M / r^2, which we solve for v to get
• v = sqrt( G M / r).
**
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I just cannot seem to put it all together.
------------------------------------------------
Self-critique Rating:
#*&!#*&!
#$&*
course PHY 121
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
028. `query 28
PRELIMINARY STUDENT QUESTION:
Again I am still having trouble with what equations to use...for PE change we would use the equation PE = Gmm/r1 -
Gmm/r2 same as the Force equation or do we just find dPE by grav. Force * ds both are listed in my notes. As far as
KE I assume m*9.8m/s/s *rE/r^2
INSTRUCTOR RESPONSE
PE = -G M m / r^2 so going from radius r1 to radius r2 we would have `dPE = GMm/r1 - GMm/r2.
If r1 and r2 don't differ by much, then G M m / r1 and G M m / r2 won't differ by much, and it would be important to use an appropriate number of significant figures in calculating the difference.
Alternatively the difference can be expressed in terms of the common denominator r1 * r2 as G M m * ( r2 - r1) / (r2 * r1), where the common factor G M m has been factored out.
As another alternative if r1 and r2 are proportionally close, you can figure out the average force and multiply by `dr = r2 - r1, as described below:
`dPE, i.e., the change in gravitational PE, is also equal to average gravitational force * change in distance from planet center:
• `dPE = F_ave * `dr
This is valid if `dr is small compared to r1 and r2, for the following reasons and with the associated cautions:
Gravitational force is not linearly related to r (i.e., the graph of F vs. r is not a straight line), so the average force on an interval is not equal to the average of the initial and final force.
However if the proportional change in r is small enough the curvature of the graph won't be noticeable, and the average force could be approximated by the average of initial and final force. In fact it is possible that the proportional change in r is small enough that over the relevant interval there is no significant change in the force itself.
• In these cases the change in PE is more easily found using F_ave * `dr.
*********************************************
Question: `qQuery class notes #26
Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
`a** The proportionality is accel = k / r^2. When r = rE, accel = 9.8 m/s^2 so
9.8 m/s^2 = k / rE^2.
Thus k = 9.8 m/s^2 * rE^2, and the proportionality can now be written
accel = [ 9.8 m/s^2 * (rE)^2 ] / r^2. Rearranging this gives us
accel = 9.8 m/s^2 ( rE / r ) ^2, which we symbolize using g = 9.8 m/s^2 as
a = g rE^2 / r^2.
If we set the acceleration equal to v^2 / r, we obtain
v^2 / r = g ( rE / r)^2 so that
v^2 = g ( rE^2 / r) and
v = sqrt( g rE^2 / r) = rE sqrt( g / r)
Thus if we know the radius of the Earth and the acceleration of gravity at the surface we can calculate orbital velocities without knowing the universal gravitational constant G or the mass of the Earth.
If we do know G and the mass of the Earth, we can proceed as follows:
The gravitational force on mass m at distance r from the center of the Earth is
F = G m M / r^2,
Where M is the mass of the Earth and m the mass of the satellite. Setting this equal to the centripetal force m v^2 / r on the satellite we have
m v^2 / r = G m M / r^2, which we solve for v to get
• v = sqrt( G M / r).
**
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I just cannot seem to put it all together.
------------------------------------------------
Self-critique Rating:
#*&!#*&!#*&!
@&
It's best to know this, but as long as you know
F = G * m1 * m2 / r^2
you can do OK here.
*@