The Simulation Gravity and Orbits - Assignment Example

4. Can you explain how your previous choice makes physical sense? The earth is constantly circularly rotating the sun. Since the position of the earth from the sun is fixed at a constant radius, there is an indication that the forces of gravity of the earth and the sun are equal and constant. Go to Earth-Sun in the "to scale" mode. Click to show the masses of each object. At this point, the Earth's period should be 365 days (if it is not, just hit the reset button). Double the Sun’s mass. What is approximately the new period? Dont forget your units!504 Thousand Earth Masses6.

Hit reset to go back to 365 days. Keep the Sun’s mass at its original value but now double the mass of the Earth. What is the new orbiting period? 1.51 Earth Masses7.From Newton’s law of gravity, we can work out the following formula for the orbital velocity for the circular motion:v2 = G where v is the orbiting speed, r is the orbital distance and G is Newtons constant (see the course page for more on this). The orbital period is just the distance traveled in one revolution divided by the speed.

So the period T is inversely proportional to the velocity (T~1/v).In the formula for the velocity (or period), the mass M is the mass of which object?Earth8. According to Newton's theory of gravity, if I double the mass of the Sun, by how much should the velocity of the Earth change? How much should the period change? Does this agree with what you found in question 5? If not, why not? If the mass is doubled and the radius is kept constant, the velocity square of the velocity will double. This agrees with the earlier findings.9. If I double the radial distance between the Earth and the Sun (keeping the orbit circular) what happens to the orbital speed? A) It increases by the square root of 2 (about 1.4)B) It decreases by the square root of 2 (about 1.4)C) It doubles) It is reduced by 2B) It decreases by the square root of 2 (about 1.4)10. Now imagine that instead of orbiting around an object you are orbiting inside of an object of uniform density (this is possible if you are orbiting inside of a thin gas with no drag for example).

We will assume the orbit is circular. Newton showed that the only mass that matters for your orbiting motion is the mass inside of the radius you are orbiting as the figure below shows. where V is the volume which goes like the radius cube. The mass inside of a radius r increases as the volume increases. As I move the satellite further away from the center, there is more and more mass inside the orbit. If double the radius (still saying inside the gas), what happens to the velocity? v2 = G M/rHint: Newton's formula for v is good here but the mass affecting the orbital motion now increases as you move out (see figure). A) The orbiting velocity increases by the square root of 2 (about 1.4)B) The orbiting velocity decreases by the square root of 2 (about 1.4)C) The orbiting velocity doubles) The orbiting velocity is reduced by 2B) The orbiting velocity decreases by the square root of 2 (about 1.4)11. The figure below shows the measured speed of stars (in km/s) orbiting a neighbor galaxy as a function of distance from the center of that galaxy.

Kpc stands for kilo parsec and it is an astronomical unit to measure distance. In the figure, we added the image of the galaxy. Almost all the visible matter ends at 4 kiloparsecs. Can you explain the graph between 1 kpc and 4 kpc? What do you think is going on between 4 kpcs and 6 kpcs? Between 1kpc and 4Kpc As the distance (Radius) from the galaxy increases the speed of orbit decreases. Between 4KPC and 6 KPc – Despite an increase in the radius increases the velocity is increasing.