(2) At exactly the same time that you drop a valuable crystal glass, your housemate throws a bottle horizontally. The bottle lands at the feet of your noisy neighbor--which smashes first--the glass or the bottle? (Hint: think of the horizontal and vertial motions separately). The VERTICAL motion is entirely controlled by gravity - it is only the horizontal motion which is affected by the throw (assuming it is thrown exactly horizontally with no upward or downward component). So, this means that just like when we drop 2 objects they hit the ground at the same time, then dropping and throwning horizontal should produce the 2 objects hitting the ground at the same time.

(3) If you drop an object it lands on your toes. If you throw it sideways it goes a way before hitting the ground. If you throw it harder sideways it goes farther. If you throw it up at an angle does it go farther? Always? (Remember: if you throw it directly up in the air it lands on your head) If you throw the object upwards at an angle, it goes up a way before starting to fall down. So, the object spends more time in the air and allows your horizontal force to act for longer - so it goes farther. BUT - if you throw it at a steeper and steeper angle, at SOME POINT you are not adding much horizontal force, mostly upward force - so the object does not end up as far away (in the limit of no horizontal force it lands on your head!) To maximize the distance you need to throw upwards at an angle of 45 degrees. Try it!

(4a) Why are satellites generally launched towards the East? (Even after the end of the Cold War.) Hint: Think about the Earth's spin. Remember, NY leads LA - so the Earth spins to the East. If you launch a rocket towards the east, the rocket already has the spin motion of the Earth - makes it easier to get up to orbital speed

(b) Why are launch facilities nearly all built on the east coasts of continents? If you are launching to the East, then you want the rocket to go into the ocean if it does not make it and blow up!

(5) The Hubble Space Telescope (mass=2000 kg) is in orbit at about 580km above the Earth. Also orbiting the Earth at 580 km is one of the astronauts gloves (mass=0.5 kg). Which orbits the Earth faster, HST or the glove? The HST and the glove orbit the Earth at the SAME orbital speed - the formula for the speed of an orbiter does NOT have the MASS of the ORBITER in it (just the mass of the orbitEE - in this case, the mass of the Earth).

(6) Above are diagrams showing orbits when launched with increasing speed. Label them appropriately with the terms: A=Circular, B=elliptical, C=Escape, D=Sub-orbital (1) is D=sub-orbital, (2) is A=Circular, (3) is B=Elliptical, (4) is C = Escape.

(7) In a circular orbit 200 m above the surface of the Earth (that is 6400+200=6600 km from the center of the Earth)--there is a spy satellite looking down. We can work out the orbital period of this spy satellite (and, obviously, so could the people being spied on)--as follows:

(a) What is the total circumference of the orbit?

Circumference = 2 radius = 2 6600km = 41,000 km

(b) Remember SPEED = DISTANCE / TIME, so TIME = Distance / Speed

(c) If the DISTANCE is in km, and the SPEED is in km/s, what are the units of TIME? Km / km/s = s (i.e. seconds)

(d) Now plug in the numbers and convert the orbital period of this spy satellite to hours. The spy satellite is orbiting 41,000 km at a speed of 8 km/s = 41,000/8 s = 5125 s = 5125/3600 hours = 1.4 hours (85 minutes).

(8) Check that it still works for Jupiter where a_J = 5.3 A.U.--what is P_J (in years)?

P²=a³
P= (a)^3/2
So, P_J = (a_J)^3/2=(5.3)^3/2=12.2 YEARS

(9) So, the value of ( 4 ² / G ) = 4 x (3.14)² / (6.67 x 10^-11) = 5.9 x 10¹¹

Thus we have P² = 5.9 x 10¹¹ x (a³ / M) with P in seconds, M in kg and a in meters

(10) Write out the equation for Newton's Version of Kepler's Third Law, re-arranging the terms with each of the 3 variables on the left hand side of the equation - as if you wanted to solve for

(a) P = [5.9 x 10¹¹ x (a³/M]^1/2

(b) a = [P² M /5.9 x 10¹¹]^1/3

(c) M = 5.9 x 10¹¹ [a³/P²]

(11) Say we measure the orbital period, P, and the orbital distance, a, of one of the moons, say Europa, orbiting Jupiter. What can we determine from this equation?

(a) the mass of Europa. NO - Europa is the orbitER

(b) the mass of Jupiter. YES! Jupiter is the orbitEE

(c) the mass of the Earth. Nah - Earth's got nothing to do with Europa orbiting Jupiter.

(12) Stars often come in twos and threes rather than on their own. Say we detect a little star orbiting a big star. The little star has an orbital period around the big star of about one year. The distance between stars is harder to measure, but say we estimate it to be about 1 A.U. What is the mass of the big star compared to our Sun? Remember, we measure the orbital period and orbital distance of the orbitER and that it is the orbitEE's mass that counts, not the orbitER. In this case, the orbital period and orbital distance of the small star are the SAME AS THE EARTH. So, the mass of the orbitEEs must be the SAME - the big star has the same mass as our Sun.

(13) We can "weigh" a galaxy by measuring the speed of stars at its edge. Describe how you think this might be done using Newton's Version of Kepler's Third Law. The orbitEE is the galaxy. The orbitERs are stars on the edges of the galaxy. So, we measure the orbital period and orbital distance of the stars at the edge of the galaxy and this tells us the mass of the galaxy.