Slingshot effect and g force (1 Viewer)

Wohzazz

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Can someone explain to me in simple terms the slingshot effect, like how it works, and why space probes exploit it.

Also how do what is g force and how do you calculate it? I know its g force= g + a/ 9.8
do we always assign a as positive or does it depend on which directions it travelling?
and g is always postive right?
 

Constip8edSkunk

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Basically as the space probe swing around a massive body, the gravitational field of the planet accelerates the probe... conversely the planet slows down slightly... this can b see by the conservation of momentum (probe's mass is very small so it speeds up alot while the planet's mass is huge so the velocity change is negligible)

Using a simple example:
Assume a frame in which say jupiter is stationary, a space probe is travelling towards jupiter at 28000 ms^-1, it 'rebounds'(which isnt actually what happens, but lets assume a 180 degree change) away elastically at 28000ms^-1. In this frame the sun is also travelling towards jupiter at 13000ms^-1 so the speed of probe relative 2 the sun becomes 41000ms^-1



edit: a doesnt have to be positive, it depends on the direction of the acceleration relative to the gravitational acceleration. as for the signs... it depends on which direction you assign as being positive. the directions are always relative ... it wouldnt affect the end result
 
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Wohzazz

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first of all, i'm not really familiar with the word 'frame' in physic context
and as for the example, is the space probe travelling at 28000ms^-1 relative to jupiter or the sun? are they travelling head on?
 

Wohzazz

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for g force, i saw this example that if you free fall in a life you feel weightlessness meaning you feel zero g force
so g for this is assigned +9.8 while a is assigned -9.8, this works you but i don't get why
 

Constip8edSkunk

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the example isnt realistic, its just easier 2 explain heheh

frame as in frame of reference.

probe is travelling towards jupiter head on at 28k

in your example, a represents air resistance which is in the opposite direction to g

edit: the button is on the bottom right corner of the post... lift?
 

Wohzazz

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well if its head on, doesn't it mean that the relative motion of jupiter and the space probe is 28000ms^-1, doesn't mean when you add it up, you get the speed relative to the sun....its all this relative stuff that i don't get

and for a equaling air resistance, i thought it was acceleration due to gravity, why would air resistance have a 9.8ms^2?
 

Wohzazz

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Lift instead of 'life' in the post previously
And serious my edit button is gone, i use i know it's at the button usually, but its not here now
 

Xayma

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Originally posted by Wohzazz
Lift instead of 'life' in the post previously
And serious my edit button is gone, i use i know it's at the button usually, but its not here now
Witht he forum upgrade the quick edit and quick quote buttons are currently MIA. In the bottom right of your post you will see two buttons "edit" and "quote". Quote brings you to a reply page with the text quoted. Edit allows you to edit your post.
 

Constip8edSkunk

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probe relative to jupiter :
1st 28k towards then 28k from
sun to jupiter:
13k towards

so probe to sun, 41k towards


the acceleration due to gravity is always considered 9.8ms^-2 over such short distances to earths surface.... the acceleration due 2 air resistance opposes the direction of motion and increases as the velocity increases so at some point (terminal velocity) it balances out with gravitational acceleration, so u experience freefall

yeah i was visualising some1 skydiving rather than a lift, but the same thing i suppose
 

Xayma

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Originally posted by Wohzazz
for g force, i saw this example that if you free fall in a life you feel weightlessness meaning you feel zero g force
so g for this is assigned +9.8 while a is assigned -9.8, this works you but i don't get why
-9.8ms^-2 would be the decelleration applied by the Fricitonal Force of the air (with the 9.8 by gravity cancelling out with it) resulting in you moving at a constant velocity.
 

Wohzazz

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Originally posted by Constip8edSkunk
probe relative to jupiter :
1st 28k towards then 28k from
sun to jupiter:
13k towards

so probe to sun, 41k towards
so your saying jupiter is initially travelling towards the sun and sun towards it
shouldn't probe after sling 'rebound' be less than 28k ms^-1 because they are travelling in same direction
how about for calculations of g force on astronaut while a spaceshuttle is re-entering earth atmosphere and is decelerating
g force =g+a/9.8
g=9.8
a=?
 
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Constip8edSkunk

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O..................*.....o
Sun..........probe..Jupiter(assume stationary)

-->13k.........----> 28k
then:
-->13k .......<---- 28k

so final velocity = 13k+28k = 41k
 

Wohzazz

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ok i get it now:)
Skunk, can you please answer my question 2 posts above this one?
 

Constip8edSkunk

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assuming the engines off, you have 2 forces acting on the shuttle:

^ air resistance(ma)
|
SHUTTLE
|
v gravity(mg)

note air resistance is opposite in direction 2 gravity, so opposite signs
 

Wohzazz

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the book says you could feel 20g going down(those traditional shuttles) How is that possible? the thrusters must be on?
 

zeropoint

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Originally posted by Wohzazz
the book says you could feel 20g going down(those traditional shuttles) How is that possible? the thrusters must be on?
I fail to understand this conclusion. Using Newton's second law, the net force acting on the spacecraft is

F = mg + F<sub>fric</sub>

where g is the gravitational field vector and F<sub>fric</sub> is the frictional force exerted by the air molecules. Resolving into perpendicular components

F<sub>x</sub> = −F<sub>fric</sub> cos(theta) i

F<sub>y</sub> = F<sub>fric</sub> sin(theta) j + mg

where theta is the angle of depression of the spacecraft as it enters the atmosphere, i and j are unit vectors lying in the positive x and y direction respectively. I assume here that the spacecraft enters the atmosphere with some positive horizontal component of velocity. F<sub>fric</sub> is typically proportional to v or v<sup>2</sup>. Therefore a sufficient re-entry velocity will in principle allow any vertical acceleration, such as 20 g for instance.
 

Wohzazz

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sorry zero point but i don't know what you proved with the equations...i cannot establish a relation from velocity and g force...it doesn't matter how far, your are travelling, g force depends on acceleration and gravitational force at that point right? maybe you could be a little more explicit?
 

Constip8edSkunk

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there is no limit on the magnitude of a(because there is no limit on teh velocity upon entry) so there is no limit on the magnitude of the g-force
 

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