Friday

January 30, 2015

January 30, 2015

Posted by **Sandhya** on Wednesday, April 22, 2009 at 10:35pm.

A) Calculate the year of the planet X.

B) Calculate the acceleration due to gravity on planet X.

C) Calculate the gravitational field of the sun (acceleration due to the sun’s gravity) at Planet X.

2. A horizontal spring is compressed a distance 8cm from is equilibrium position and released against a ball of mass 50 g which is propelled away from the spring at 5 ms-1.

A) Calculate the spring constant.

B) Calculate the work in compressing the spring.

- Physics -
**TchrWill**, Thursday, April 23, 2009 at 6:24pmThere are numerous ways for determining the characteristics of orbiting bodies and the whole subject, from the historical point of view, is a facinating one.

If you know the mean distance and orbital period of a satellite orbiting the earth, such as the moon, and the mean distance and orbital period of a satellite orbiting a "new" planet, for instance, you can determine the value of gravity on the new planet, as well as all the other vital characteristics about the planet. Lets assume, for the

sake of your specific question, that Mars was just discovered and sufficient time has passed, such that astronomical observors have determined the mean orbital distance of Deimos, the outermost moon of Mars, and its mean orbital period about the planet. Its mean orbital distance happens to be 14,597 miles and its orbital period is 30.3 hours or 1.2625 days.

We already have at our disposal the mean distance of the moon from earth as 238,867 miles and its orbital period of 27.32 days. We also know the mass of the earth as 1.32x10^25 lb. and the Universal Gravitational Constant, G = 1.0664x10^-9 ft.^3/lb.sec.^2 which yields a Gravitational Constant, mu(e), for the Earth of 1.40766x10^16 ft.^3/sec.^2.

Now, from orbital mechanics, we know that the orbital period of a satellite about a body with a central gravitational field is given by T^2 = 4(Pi)^2a^3/mu where T is the orbital period in seconds, a is the mean distance of the satellite from the center of the body in feet, and mu is the Gravitational Constant of the body. Now, let T(m) be the orbital period of the moon, a(m) be the mean distance of the moon from earth, m(e) be the mass of the earth, T(D) be the orbital period of Deimos, a(D) be the mean distance of Deimos from Mars, and m(M) be the mass of Mars.

We now have T(m)^2 = [4(Pi)^2a(m)^3]/Gm(e).---(1)

We also have T(D)^2 = [4(Pi)^2a(D)^3]/Gm(M).---(2)

Dividing (1) by (2) and solving for m(e)/m(M) = [T(m)^2(a(D)^3)]/[T(D)^2(a(m)^3].

Knowing T(m) = 27.32 days, T(D) = 1.2625 days, a(D) = 14,597 miles, and a(m) = 238.867 miles, we can now determine the ratio of the mass of Mars to the mass of the earth as m(M)/m(e) = .10685. The weight of the earth is ~1.32x10^25 lbs and its mass is 4.10269x10^23 lb.sec.^2/ft. Therefore the mass and weight of Mars is .10685

times the mass and weight of the earth or m(M) = 4.3837x10^22 lb.sec.^2/ft. and W(M) = 1.4104x10^24 lb. Hence the gravitational constant for Mars becomes mu(M) = 1.4104x10^24(1.0664^-9) = 1.50845x10^15 ft.^3/sec.^2.

Now one way of determining the value of gravity, g(M) on Mars is to compute the weight of a body of mass 10 lb.sec.^2/ft. which would weigh 321.74 lb. on earth from F = W = mu(M)m/r^2. Letting mu(M) be the gravitational constant for Mars = 1.50845x10^15, m = our reference mass of 10 lb.sec.^2/ft., and r = the mean radius of Mars, which was determined by way of another clever method, and is equal to 2120 miles. Therefore, the force of attraction of Mars on our reference mass, or its weight on Mars, is W = 1.50845x10^15(10)/[2120(5280)]^2 = 120.390 lb. The ratio of its weight on Mars to its weight on earth is 120.390/321.74 = .37418. Therefore, gravity, g, on Mars is .37418(32.174) = 12.039 fps^2, which can be confirmed in any astronomical reference book.

To summarize, there definitely are ways of determining the distance, diameter, orbital period, mass, average density, and gravity of a new found planet orbiting the sun. Needless to say, it would be extremely difficult to determine these same characteristics accurately for a planet in another solar system due to the distances

involved. Of course, after reading this, you might logically ask, "How was the mass of the earth first determined, how was the mass of the sun determined, how were the distances to the planets and sun determined, how was the magnitude of the Universal Gravitational Constant determined, and so on into the night?" As I said earlier, the whole subject of the evolutionary determination of the planetary characteristics within our solar system is a facinating one. If you are at all interested in the subject, I refer you to some excellant historical books on the subject.

1--An Elementary Survey of Celestial Mechanics by Y. Ryabov, Dover Publications, Inc., 1961.

2--A History of Astronomy by A. Pannekoek, Dover Publications, Inc., 1989.

3--A History of Astronomy from Thales to Kepler by J.L.E Dreyer, Dover Publications, Inc., 1953.

4--Theories of the World from Antiquity to the Copernican Revolution by Michael J. Crowe, Dover Publications,

5--Inc.,1990.

6--Modern Theories of the Universe by Michael J. Crowe, Dover Publications, Inc.,1994.

7--Greek Astronomy by Sir Thomas L. Heath, Dover Publications, Inc.,1991.

8--The Measure of the Universe by J.D. North, Dover Publications, Inc.

9--From Copernicus to Einstein by Hans Reichenbach, Dover Publications, Inc.

10--Astronomy-A Self Teaching Guide by Dinah L. Moche, John Wiley & Sons, Inc., 1993.

11--Ancient Astronomers by Jeremy A. Sabloff, Smithsonian Books, 1993.

See if you now solve your problem.

**Answer this Question**

**Related Questions**

physics - The mean distance of the planet Neptune from the Sun is 30.05 times ...

physics help 3!! ** - Based on the following data about planet X (which orbits ...

physics - The physics of orbital radius and orbital period of a planet. If a ...

Physics - The mass of a certain planet is six times the mass of the earth and ...

physics - A planet has a radius of about 0.85 times Earth's radius and a mass ...

earth - The Earth orbits the Sun once every 365 days. The distance between the ...

physics - person weighs 526 N on Earth. Another planet has twice the mass of ...

physics - What is the ratio of the spin angular momentum of the earth and its ...

Physics please HELP - What is the ratio of the spin angular momentum of the ...

Physics - Astronomers can calculate the mass of a planet by using the orbital ...