Question

Help please

Answer 1

@Vocaloid

Answer 2

it would be right by the sun right

Answer 3

for a) yes, for b) it would be the opposite (farthest away from the sun)

still figuring out c

Answer 4

(this part has always confused me about orbits, but the potential energy is highest where the satellite is closest to the central body, then decreases as it moves farther away, then increases again, so your graph should be a parabola/U-shape I believe)

Answer 5

I'm not quite sure what I'm looking at here

Answer 6

OH the two dots

Answer 7

i labeled them just wanted to make sure the direction is right

Answer 8

like where i placed them

Answer 9

if you're labelling the closest/farthest points away on the orbit it should be at the two ends like this

Answer 10

[your system is the Earth + sun but the principle is the same]

Answer 11

so where the two arrows are that's where?

Answer 12

where the moon is not the arrows

Answer 13

yeah something like that

Answer 14

OKay now for b

Answer 15

I mean c

Answer 16

for b it's the farthest point from the central body (so the point on the far right)

Answer 17

make sure you're labelling them as "highest KE" and "lowest acceleration" since that's what they're asking for a + b

Answer 18

c should be an upward facing parabola

Answer 19

sort of like that but a little broader, plus gpe is never negative or 0 so it shouldn't touch the x-axis

Answer 20

[just make it a bit more symmetrical than that]

Answer 21

better but it could be more symmetrical

Answer 22

eh, could be a bit more even on both sides but let's move on

Answer 23

Okay sure

Answer 24

actually @sillybilly123 would the potential energy during an orbit just be a parabola or is it more complex than that

Answer 25

we can move on in the meantime

Answer 26

okay sure that is it for this question right

Answer 27

yeah pretty much

Answer 28

given \(T + U = \text{const.}\), ie we can call it a closed frictionless system, the energy equation should be:

\(\dfrac{1}{2} m v^2 - \dfrac{\mu m}{r} = \text{const.} \)

where r is distance between star (at ellipse focus) and planet, \(\mu = GM\) and v is planet speed relative to star

\(T \propto \frac{1}{r}\), most T means smallest r, ie when they are closest. ( "closest" isn't necessarily right at the perigee but is pretty close from what i can calculate)

the gravitational force is always inward toward star, and \(F = m a = - \nabla U = \dfrac{\mu m}{r^2}\) ie \(a \propto \frac{1}{r}\), so that too should be greatest for small r, ie when closest. hence the slingshot effect

graphing: the energy has to be negative, and one could guess it looks something like this just by smoothing out between 3 points:



behind that question is a load of great original math from superstars like Kepler, Newton...

Answer 29

so "lowest" acceleration when furthest away....must learn to read

and graphs look pretty similar except for axis which is that slightly nebulous -ve energy point :(

Answer 30

for Sun-Earth

https://www.desmos.com/calculator/cbq3i2bzsx