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Virtual Learning
Physics
Determining “G”
May 11, 2020
Physics
Determining “G” : May 11,
Objective/Learning Target:
Students will use a computer simulation and graphing techniques
to determine the gravitational constant “G”
Quick Review #1 Answer
The forces of attraction
ranked from greatest to
least.
B=C, A, D
Quick Review #
Suppose that an apple at the top of a tree
is pulled by Earth’s gravity with a force of
1 N. If the tree were twice as tall, would
the force of gravity on the apple be only ¼
as strong?
Newton’s Law of Gravitation and “G” Introduction You will use a computer simulation today to reinforce your ideas of Newton’s Universal Law of Gravitation and to determine “G”. Since this is “inquiry based”, you’re not supposed to know everything going in, but learn as we walk through the lesson. You must read the following slides carefully. Let’s get started!
Newton’s Law of Gravitation and “G”
The simulation will let you vary distances and masses, and
then shows the forces between them. Leaving “G” as the only
unknown.
The force of gravity between two normal-sized objects is
much too small to measure even in most college laboratories,
hence this is a good simulation.
F = G mM r²
Newton’s Law of Gravitation and “G”
Website:Gravitational Force Simulation
Make sure to use the HTML5 version.
Select Download to get started..
Newton’s Law of Gravitation and “G”
- Vary the masses and distance between the masses, record all your data along with the gravitational force that is given in the Sim. Record all sig figs.
- Do this for ten different scenarios recording each in the data table. Make sure to use the entire range possible for all the parameters. The ruler is movable so you can get the distances more exactly. This will be your least accurate measurement. Make it as carefully as you can. M 1 (kg) M 2 (kg) r (m) F (N) X = M 1 M 2 r²
Newton’s Law of Gravitation and “G”
4. Plot a F vs. “X” graph and
calculate the slope.
5. The slope is your
experimental value for ”G”.
Calculate the percent error.
Actual value:
G = 6.673 x 10⁻¹¹Nm²/kg²
F (N) X (kg²/m²)
Simulation example answers
- M 1 (kg) M 2 (kg) r (m) F (N) X = M 1 M
- 100 400 4.0 1.66852 x 10⁻⁷ r²
- 200 400 4.0 3.33704 x 10⁻⁷
- 800 400 4.0 1.334816 x 10⁻⁶
- 800 1000 2.4 9.269556 x 10-⁻⁶ 138888.
- 800 1000 6.0 1.483129 x 10⁻⁶ 22222.
- 800 1000 9.6 5.79347 x 10⁻⁷ 8680.
- 10 10 9.6 7.24 x 10⁻¹¹ 1.
- 50 500 7.0 3.4051 x 10⁻⁸ 510.
- 60 120 8.0 7.508 x 10⁻⁹ 112.
- 350 80 8.9 2.3592 x 10⁻⁸ 353.
Simulation example answers
4. Slope = Rise = (y 2 -y 1 )
Run (x 2 -x 1 )
= (1.66853 x 10⁻⁷ - 7.508 x 10⁻⁹)N
(2500-112.5)kg²/m²
=6.67409424 x 10⁻¹¹Nm²/kg²
5. %err = (experimental value - accepted value) x 100%
accepted value
=6.674 x 10⁻¹¹ - 6.673 x 10⁻¹¹ x 100% = 0.1 %
6.673 x 10⁻¹¹
Additional Practice
Return to the simulation and practice using Newton’s Law of
Gravitation by picking mass values and a distance and
calculating the Force between them using the formula and
checking your answer with the simulation.
F = G mM r²