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Projectiles in space: Calculating initial velocity

  1. Feb 8, 2016 #1
    This question is on space and projectiles. It is about comet 67P/ Churyumov-Gerasimenko and the Rosetta/ Philae lander.
    Question: 67P has a gravitational field strength of 5.2x10^5 N/Kg. After Philae bounced it took 1 hour 50 minutes to reland. Calculate the initial velocity of Philae. Does this seem to be a realistic value?

    You are allowed to get other information that you need off the internet to answer the question however, I do not know how the gravitational field strength comes into play.

    I'm preety sure you use
    u = Initial velocity,
    v = Final Velocity,
    t = time taken,
    s = distance traveled or displacement,
    a = acceleration.

    The time taken is 6600 secs. But i still need to find the acceleration, distance traveled and final velocity. I'm sure we cannot just use the distance traveled off the internet because the next question asks us to find it. I guess i can find the force applied to Philae using F=mg but i'm not sure how that would help me but it uses the gravitational field strength which is given?
     
    Last edited: Feb 8, 2016
  2. jcsd
  3. Feb 8, 2016 #2
    I found acceleration using F=ma, F/m=a
    5.2x10^-5 / 100
    (weight of Philae=100kg)

    This equaled 0.00000052m/s^2
    I then substituted into the formula u = v-at
    and got -0.003432 m/s
    This doesn't seem right as I should be getting close to 0.3m/s.
     
  4. Feb 8, 2016 #3

    SteamKing

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    On earth's surface, the gravitational field strength is 9.81 N / kg. Compare this to the field strength on 67P of 5.2×10-5 N/kg to find g for 67P.

    The mass (not the weight of Philae) is not required to find the initial velocity, once you know g for 67P.
     
  5. Feb 8, 2016 #4
    So is this on the right track?
    u = v-at
    u = 0 + 5.2x10^-5 x 6600
    u = 0.34m/s
     
  6. Feb 8, 2016 #5

    SteamKing

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    No.
    Remember, 6600 sec is the total time it took Philae to reland on the surface of 67P after it bounced.
    What is the velocity of Philae at the highest point of its travel above the surface of 67P?
    Which variable, u or v, represents the initial velocity you are looking for?
     
  7. Feb 8, 2016 #6
    So does the time just get halved because I am only calculating till its about to start falling back to comet 67p. And the velocity of Philae at the highest point of travel would be 0 m/s wouldn't it because its about to start coming back ( doesn't that make the final velocity 0?)
    And the u variable represents initial velocity? As from the question i'm assuming we need to find the velocity as soon as the bounce began.
     
  8. Feb 8, 2016 #7

    SteamKing

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    Yes, as long as the excursion which Philae takes is un-powered. It is reasonable to assume that the trip took 3300 sec. until the velocity of Philae relative to 67P reached zero, after which the gravity of 67P pulled Philae back to its surface.
    Yes, at the highest point above the surface of 67P, the velocity of Philae is zero.

    You don't care what the final velocity is coming back down, because you are not asked that.

    BTW, when Philae hits the surface of 67P, its velocity will not be zero.
    Look at the equation:
    u = v - at

    u only occurs after some time t has elapsed, so it must be the velocity t seconds after the velocity v has occurred. The acceleration a is acting to reduce the velocity v to give velocity u.

    In your calculations above, the minus sign in the equation magically transformed into a + sign. You can't arbitrarily switch signs like that in order to make the answers come out right and still obey the formula.
     
  9. Feb 8, 2016 #8
    I don't completely understand what you are trying to explain.
    I did
    u = v-at
    u = 0 - - (+) 5.2x10^-5 x 3300
    u = 0.17 m/s
    (the 5.2x10^-5 is negative because its going up, similar to an object on earth which would be -9.8m/s upwards and 9.8m/s downwards?)

    Sorry for not understanding the point you are trying to make, but does that seem like the right answer? (in reality it is the wrong answer however but i think the teacher purposefully gave Comet 67p too much gravitational field strength)
     
  10. Feb 8, 2016 #9

    haruspex

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    Maxie's work looks ok to me there. u is the velocity just after bounce, v is the velocity at top of bounce, a is the acceleration. V=u+at. With up positive, a is negative, v is 0, and u will turn out positive.
     
  11. Feb 8, 2016 #10
    I then have to calculate how high Philae is likely to have bounced.
    I used this formula:
    vertical_displacement_equation.png
    or delta Y = ut + 1/2at^2
    and i got 277.86m
    The actual recorded value was around 1000m so do either of you think this was on purpose or am I incorrect?
     
  12. Feb 8, 2016 #11

    haruspex

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    The 5.2*10-5N/kg seems low. Net searches say ten times that.
     
  13. Feb 8, 2016 #12

    SteamKing

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    Yes, but on earth as on 67P, if something has a velocity leaving the surface, the common convention is to treat that velocity as positive, and since the only change in velocity comes from gravitational acceleration directed back toward the center of the earth (or, in this case, 67P), then the initial velocity is being reduced by the quantity ##at## until u = 0, after which, the satellite falls back to the surface.
    Again, there is a minus sign in the original formula which magically converts to a plus sign for no reason.

    The minus sign is there in the original equation because the object is not going to travel a distance of vy0 ⋅ t, since the upward velocity is being continually reduced by the acceleration due to gravity, until the velocity reaches zero.

    The estimated escape velocity for 67P is also quite low, only 1 m/s. You could literally jump off this rock if you were standing on it. :wink:

    https://en.wikipedia.org/wiki/67P/Churyumov–Gerasimenko
     
  14. Feb 9, 2016 #13

    haruspex

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    No, until v=0. As I posted, Maxie is using u for the bounce velocity.

    That same site quotes https://en.wikipedia.org/w/index.php?title=Template:5_x_10%5E-4&action=edit&redlink=1 [Broken] m/s2
     
    Last edited by a moderator: May 7, 2017
  15. Feb 9, 2016 #14
    I'm confused as to what the right answer is now. SteamKing, can you show me how you would work out the question? If I use a minus sign in my calculator when calculating the distance of the bounce I get a syntax error? and i understand what you mean about having velocity as positive but its just what we have been taught to do in year 12 Physics...
     
  16. Feb 9, 2016 #15

    SteamKing

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    If you evaluate the equation

    $$Δy = v_{y0} ⋅ t - \frac{1}{2}gt^2$$

    for t = 3300 sec; g = 5.2 × 10-5 m/s2; and vy0 = 0.17 m/s

    What possible syntax error would you get? It's a straight forward plug and chug. You shouldn't be doing anything illegal, like taking the square root of a negative number ...
    I think that's the problem. u should be the velocity at time t, given that v is the initial velocity.
    That's the way I interpret the formula.
     
  17. Feb 9, 2016 #16
    Sorry I have no idea why I was typing + instead of -. All along I was using minus. So i got the answer of 277.86m. Even though thats very far off the real value I guess thats what the teacher wants us to realise with the "Does this seem to be a realistic value?". I guess it is still a realistic value as its lower than 1m/s which is Comet 67p's escape velocity.
    Thanks for all the help though! Hopefully its correct. I'll be sure to use this website again in the future :)
     
  18. Feb 9, 2016 #17

    haruspex

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    I don't know why you are interpreting it that way. If not using the vi, vf notation, a common form is u for initial, v for final. All Maxie did was take the usual form of that, v=u+at, and turn it round into u=v-at. If you read Maxie's posts in that light it all makes sense.
     
    Last edited: Feb 9, 2016
  19. Feb 9, 2016 #18

    haruspex

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    As I wrote in post #11, the reason your answer does not match experience is that the value they gave you for gravitational acceleration at 67P is out by a factor of 10. For the acceleration given, your answer is realistic.
    Note, though, that 67P is not terribly big. What is unrealistic about the method you have used?
     
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