Question

1) a satellite is in a elliptical orbit around earth. at one point on the ellipse, it is 25 x 10^6m from the center of earth and has a speed of 12,000 m/s. what is its speed when its speed is at another point on the ellipse, 95 x 10^6 m from the center of earth? the mass of earth is 5.97 x10^-11 Nm^2/kg^2.

A) 12000 m/s
B) 11000 m/s
C) 10000 m/s
D) 9000 m/s
E) 8000 m/s

Answer 1

This is my work, but it is not a choice:

mass of earth is 5.97*10^24 kg

r1=25*10^6 m
v1=12000 m/s
r2=95*10^6 m
v2=?
M=5.97*10^24 kg

centrifugal force on 2nd point be F1
F1=GMm/r2^2 [m=mass of satellite]
mv2^2/r=GMm/r2^2
v2^2=GM/r2
v2^2=6.67*10^-11*5.97*10^24/95*10^6
v2=2047.33 m/s

Answer 2

I think your approach is wrong because the centripetal force formula is for circular orbits.

I'm not quite sure how to solve it, but this may help: http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Gravity/OrbitalVelocities.html

Answer 3

Using that link, the fact that mechanical energy is constant for an orbiting body means you can solve for v2...

see the equation in the box in the middle, all i did was equate the mechanical energy when it's 25 x 10^6m from the center of earth, with mechanical energy when it's 95 x 10^6 m from the center of earth, since mechanical energy must be constant at any point in the orbit!

Notice you can factor out and cancel off m. Then just plug in all the numbers and find v2.

Answer 4

@Summersnow8

Answer 5

@agent0smith sorry I didnt see you responded. so what equation am I suppose to use?

Answer 6

Do you see the attachment? I used this: http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Gravity/Gifs/Gravity34.gif

Answer 7

Mechanical energy has to be the same no matter how far or close the orbiting object is.

Answer 8

right, but you never showed me how to solve it

Answer 9

Well i left the last part up to you :P at the bottom where i left off.. \[\Large 56072040 = \frac{ 1 }{2 }v^2 - 4.192 \times 10^6\]from here it's just algebra to solve for v.

Answer 10

Add 4.192x10^6 on both sides, then solve for v.

Answer 11

@Summersnow8 understand...?

Answer 12

@agent0smith is that correct?

Answer 13

Yes :)

Answer 14

okay, thank you! I was working on it all day

Answer 15

No prob :) Do you understand the method, using the mechanical energy being equal at both distances?

Answer 16

not really, I dont really get it, but its okay, im bad at physics

Answer 17

Oh :( But does it look at all familiar, the ME equation in the first link I posted...?

Answer 18

no :/

Answer 19

Oh... idk if you're meant to derive it or what then. I haven't seen your notes or anything, so idk what kinda equations you have used. You might want to ask your teacher about the question.

Answer 20

ok

Answer 21

Okay. Your first attempt was reasonable btw :) but it doesn't work because that method is for circular orbits, not elliptical.

Answer 22

hmm okay

Answer 23

You shouldn't put yourself down though, by saying "im bad at physics"... your first attempt shows a pretty decent understanding of centripetal force and gravitational force :)

Answer 24

well i just guessed

Answer 25

Well good guess then :P