SOLUTION: PLEASE HELP ME!! THANK YOU LOTS find the mass and radius of three of the nine planets in our solar system. Be sure that the masses are expressed in kilograms and the radii are e

Question 88295: PLEASE HELP ME!! THANK YOU LOTS
find the mass and radius of three of the nine planets in our solar system. Be sure that the masses are expressed in kilograms and the radii are expressed in meters.
Using your data, calculate the gravitational acceleration on each of the three planets you selected. Note: With the masses measured in kilograms and the radii in meters, the units of gravitational acceleration would turn out to be meters per squared seconds.
Use the gravitational accelerations that you calculated in Step 2 to find the period of a 2 meter long simple pendulum on each of the three planets. Note: The length of a simple pendulum is normally expressed in meters, so it is sufficient to replace L with the number 2 in the period expression.
Please submit your assignment.

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Here are the planetary masses and radii. Source is:
Merrill Physics Principles and Problems, Glenco, McGraw Hill 1992, ISBN 0 - 675 -02645 - 8, page 159
.
Mercury, mass 3.2 x 10^23 kg ..... radius 2.43 x 10^6 m
Venus, mass 4.88 x 10^24 kg ..... radius 6.073 x 10^6 m
Earth, mass 5.979 x 10^24 kg ..... radius 6.3713 x 10^6 m
Mars, mass 6.42 x 10^23 kg ..... radius 3.38 x 10^6 m
Jupiter, mass 1.901 x 10^27 kg ..... radius 69.8 x 10^6 m
Saturn, mass 5.68 x 10^26 kg ..... radius 58.2 x 10^6 m
Uranus, mass 8.68 x 10^25 ..... radius 23.5 x 10^6 m
Neptune, mass 1.03 x 10^26 ..... radius 22.7 x 10^6 m
.
The equation you need to find the acceleration due to gravity is derived from two of Newton's
developments.
.
Newton established the Law of Gravitation in which he stated that the gravitational
force of attraction between two masses is given by the equation:
.

.
In this equation G is the Universal Constant of Gravitation and is equal to 6.67 x 10^(-11) Nm^2/kg^2.
M is the mass in kg of one body. For this problem the subscript p identifies M as being
the Mass of the planet. m is a second mass which, for this problem, is a mass of a body "b"
located at the surface of the planet. R is the distance between the centers of the two
masses. In this case the subscript p means the distance between the center of the planet
and the center of mass of a body located at the surface of the planet.
.
Although the force of attraction between the planet and the body at its surface will cause
both the planet and the body to attempt to move toward each other, the difference
in their masses is so great, that only the body will experience a significant acceleration.
.
This equation tells us the force between the planet and the body. The body will accelerate.
And Newton told us that force = mass times acceleration. In this case the mass is the
mass of the body (m subscript b) times the acceleration, which for a falling body is
called g, the acceleration due to gravity. We can set the two forces equal to get:
.

.
Notice that m with subscript b appears as a factor on both sides of this equation.
Therefore, dividing both equations by reduces the equation to:
.

.
Notice now that we can calculate g for the planet just by plugging in the Universal
Constant of Gravitation, the Mass of the planet, and the Radius of the planet.
.
Let's do it for the Earth, because physics students know that the acceleration due to
gravity on Earth is 9.8 m/sec^2 and we can use that as a check.
.
Substitute 6.67 x 10^(-11) for G, 5.979 x 10^24 for M, and 6.3713 x 10^6 for R.
When you
make those substitutions, the equation becomes:
.

.
On the left side the numerator is found by multiplying 6.67 times 5.979 to get 39.87993 and
by multiplying 10^(-11) times 10^24 to get 10^13 (by adding the exponents of -11 and +24).
Furthermore, the denominator is found by squaring 6.3713 to get 40.5935 and by squaring
10^6 to get 10^12. Substituting these values results in:
.

.
Divide 39.87993 by 40.5935 and you get 0.98242. Divide 10^13 by 10^12 (subtract exponents)
and you get 10^1 which is normally written as just 10.
.
So the answer is 0.98242 times 10 or 9.8242 meters per second squared. That sure checks with
the value that is normally used in physics problems.
.
Next we need to find to find the period T of a simple pendulum. If we limit the displacement
of the pendulum to 20 degrees or less from the vertical, we can use the formula:
.

.
provided that the pendulum rod or string has negligible mass compared to the mass of
the body at the end of the string. Our pendulum is to have a length of 2 meters (from the
center of mass of the body to the end of the string) and, for Earth, we now have a value
of g = 9.8242 m/sec^2.
.
Substituting these values we get:
.

.
I did some rounding here in the calculations, but the result is that the period of a pendulum
here on Earth is 2.8350 seconds.
.
Hope this helps you. All you need to do now is select the mass and the corresponding
radius for some of the other planets. Plug these values into the equation and solve for
the acceleration due to gravity at the surface of that planet. Then plug that value of
gravity acceleration into the pendulum formula (along with the length of 2 meters) and
you will find the period of the pendulum.
.
Good luck ...
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