L1,L2
Name TWO assumptions made in projectile motion.
Only force is gravity, gravity is constant, etc. etc.
In circular motion:
Direction of centripetal force is ___.
BONUS: Direction of instantaneous velocity is ___.
Radially inwards (or synonym).
BONUS: Tangential (or synonym).
Asteroids in circular orbit experience uniform circular motion.
True or False?
True. No tangential forces to vary v_instantaneous.
When does K = GMm/(2r)?
When the mass is in circular orbit.
Optimal Perm (10^4 kg) unfortunately froze into ice on his way to Technotron, a 10^15 kg planet. Even more unfortunately, he misses it, and starts orbiting at 10^4 m from its centre.
What is the centripetal force on him?
F_c = F_g = GMm/r^2
= (6.67*10^-11) (10^15) (10^4) / (10^4)^2
= 6.67 N
When a projectile is launched, without being on flat ground.
What is the minimum number of given variables required to determine the entire set of flight conditions? (e.g. u, theta, v, h_max, h_init, etc.)
3.
Forces accelerate masses to change the direction and magnitude of their velocity.
For forces that stay F ⊥ v, [speed/direction/both] is changed.
For F ∥ v, [speed/direction/both] is changed.
Direction (e.g. centripetal F)
Speed (e.g. tangential F)
What happens when an asteroid has F_g = F_c ?
It is in circular orbit.
v_orb = sqrt(GM/r)
etc.
Two satellites are orbiting their respective celestial bodies at the same orbital period but different radii.
How is this possible?
r^3 = (GM)/(4pi^2) * T^2
They're orbitting celestial bodies with different masses.
The sheriff shots warning shot which travels 1.96km vertically upwards before falling back down and hitting him on the head.
What was its final velocity?
v^2 = u^2 + 2as
v^2 = 0 + 19.6 * 1960 (on the way down)
v = 196 ms^1
Describe how vertical and horizontal velocity vary over time, for a projectile.
Vertical velocity decreases linearly from the initial, passing 0 at the max height to reach a negative minimum at impact.
Horizontal velocity is constant at the initial.
Explain how changing the width of the screwdriver handle, and length of the screwdriver shaft, affects its effectiveness.
Torque = F_perp * r
Fatter handle = larger r (force arm in plane of rotation)
Longer shaft = no effect (length not in plane of rotation)
Compare the orbital velocities and periods of LEOs and GEOs.
v = sqrt(GM/r), i.e. v decr. as r incr.
r^3 = k * T^2, i.e. T incr. as r incr.
LEOs orbit lower than GEOs, so they have higher orbital velocity and lower orbital period.
State Kepler's 3 laws.
1. Orbits are elliptical, with the sun at one focus.
2. Satellites sweep out equal sectors in equal time.
3. T^2 proport. r^3.
A poorly designed rollercoaster has the minimum required speed to stay in contact to the tracks of a 19.6m high loop de loop.
What is its speed at the top of the loop?
F_net = F_g + N
At top, N=0, therefore F_net = F_c = F_g
mv^2 /r = mg
v = sqrt(gr)
v = sqrt(9.8*9.8) = 9.8ms^-1
A 11 y.o. sits at the back of a bus, which is doing circular motion on a roundabout. He tries to hit his sworn nemesis at the front with a paper ball, but his projectile misses because he doesn't account for the bus turning (he's 11).
Describe the ball's path from a perspective on the bus versus the ground.
Relative to ground, regular parabolic projectile motion, with bus's intantaneous v added on.
Relative the bus, vertical component of motion same, but on horizontal plane it would appear to have an acceleration that only changes in direction. It would curve to towards the bus wall away from the centre of rotation.
List TWO similarities and TWO differences between masses undergoing uniform circular motion on banked tracks and swinging strings.
Similarities
1) v = sqrt(grtanT)
2) F_net = F_x = F_c, i.e. F_y = 0
Differences
1) String: Tension_x supplies F_c,
B. Track: Normal_x supplies F_c
2) B. Track can have friction, giving it a range of v for which any r can be maintained
ETC ETC.
List TWO types of satellites.
Describe each type's rough orbital altitude and TWO potential use cases.
LEO: 200~2000km altitude
Telecomms, imaging, spyware, ISS
GEO: ~36000km altitude
Telecomms, broadcasting, weather, navigation, etc.
Explain why the direction of escape velocity does not matter, assuming it does not crash into anything.
Gravity only acts on v_rad, where K --> U as a_g reduces v_rad.
Gravity doesn't act on v_tang, hence no work is done on this axis.
But even if v = v_tang initially, the spherical g field means v_tang becomes v_rad as r --> inf.
Hence, the work to escape the field will always eventually be done.
The Broker operates a wrecking ball to destroy the Betmobile, which is driving around a 45deg angled bank track spanning 1960m in diameter.
What is the minimum angular velocity Broker needs to wipe out Betman?
v = sqrt(gr*tanT)
v = sqrt(9.8*980*1)
v = 98ms^-1
w = v/r
w = 98/1960
w = 1/20 = 0.05 rad/s
i.e. need more than 0.05rad/s to work.
Name ALL the other parameters you can derive from t_flight alone, assuming the ground is flat and g=9.8ms^-2.
t_maxheight
u_y
v_y
h_max
(also instantaneous v_y and s_y for any t)
A double-sided loop de loop track accommodates 2 separate carts at once, which start at the bottom of the loop with different initial velocities (u). Both centres of mass are effectively on the track, but one cart goes over the track, while the other goes under.
What is the relationship between the top cart's u_max, and the bottom's u_min to make the loop? Explain.
Top: At peak, F_c = F_g - N
N = 0 at v_max, therefore mv^2 /r = mg
i.e. v = sqrt(gr)
Bottom: At peak, F_c = F_g + N
N = 0 at v_min, therefore same, v = sqrt(gr)
By K --> U, top v_max = bottom v_min implies top u_max = bottom u_min. (centres of mass both effectively in the track)
Does the Earth or Moon have a faster orbital velocity about their respective central body? Propose and reason your answer.
(Sun mass being around 2*10^30 kg may help)
v^2 = GM/r
Therefore v = 2pi(GM/r)/T
v^3 = 2pi*GM/T
Therefore v^3 proport. M/T
i.e. larger M/T --> larger v
T for Earth's orbit is 365 times larger. (yr vs day)
M for Earth's orbit is 10^6 times larger. (Msun vs Mearth)
Therefore Earth orbit has larger M/T ratio, i.e. it orbits with greater orbital velocity.
At the apogee, the moon's F_g is too high to maintain circular orbit with its current v. At the perigee, F_g it's too low.
Explain why satellites will not simply revert to lower/higher circular orbits when v =/= v_orbital, and instead swings elliptically.
v has radial and tangential components. A circular orbit requires a constant v_tang, and v_rad = 0.
The equation F_c = (mv^2)/r describes the v_tang required for a circular orbit for that r.
Between the apogee and perigee, there is a point where v_tang satisfies what's required for circular orbit, but v_rad = 0 only at apogee and perigee, hence the condition for circular orbit cannot be required.
Millionaire Melon Musk is trying to catch up to Beff Jezos, who launched at v_esc into deep space. He realises that by going beyond v_esc, he can asymptotically approach v_f = 1km/s instead of v_f = 0.
What launch v_i from the Earth's surface does he need to satisfy this whim? (Assume no air res., etc., etc.)
U_inf + K_inf = E_inf = E_surf = U_surf + K_surf
0 + 1/2*mv_f^2 = -GMm/r_E + 1/2*mv_i^2
v_f^2 = -2GM/r_E +v_i^2
v_i = sqrt(v_f^2 + 2GM/r_E)
= sqrt[1000^2 + 2*(6.67*10^-11)*(6*10^24)/(6.371*10^6)]
= 11253 m/s
(Notice v_esc is already 11208m/s)