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heena

  • 2 years ago

why the escape velocity is independent of mass of projecting object and angle of projection?

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  1. heena
    • 2 years ago
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    @Callisto @Mani_Jha @apoorvk @Diyadiya @shivam_bhalla

  2. shivam_bhalla
    • 2 years ago
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    See \[{GM_e M_o \over R_e +h}= {1 \over 2} m_o v^2\]

  3. shivam_bhalla
    • 2 years ago
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    M_e --> Mass of earth R_e -->Radius of Eart M_o-->Mass of projectile v-->escape velocity

  4. shivam_bhalla
    • 2 years ago
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    The above equation is by energy conservation

  5. apoorvk
    • 2 years ago
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    buddhi lagao. Energy concept. the potential barrier has to be crossed. apply shivam's equation above.

  6. shivam_bhalla
    • 2 years ago
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    So M_o cancels and as @apoorvk said, buddhi lagao :P

  7. apoorvk
    • 2 years ago
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    :D

  8. heena
    • 2 years ago
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    lol.. anyways i need a theoretical ans if u guys see the formulae there is a presence of mass isntd ?? which show it dependent on mass right?

  9. shivam_bhalla
    • 2 years ago
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    R---> Radius of earth h--> height of projectile from earth

  10. shivam_bhalla
    • 2 years ago
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    @heena , "Open your eyes" It shows dependence on "MASS OF THE EARTH"

  11. apoorvk
    • 2 years ago
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    Okay you mean you want a 'feel'? Hmm. Okay.

  12. swalah
    • 2 years ago
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    shiv what is buddhi lagao ?

  13. shivam_bhalla
    • 2 years ago
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    "Use your brains " :P

  14. heena
    • 2 years ago
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    mean use ur brain grrrrrrr.........

  15. apoorvk
    • 2 years ago
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    buddhi lagao ---> translates to --> apply brains

  16. apoorvk
    • 2 years ago
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    *sic

  17. shivam_bhalla
    • 2 years ago
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    @heena , aur kuch (anything else ) ??

  18. heena
    • 2 years ago
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    ok means it is depends on mass but mass of earth right :) thnk you all of u meri khinchai krne k liye :P popat kr diya mera hur. :D

  19. shivam_bhalla
    • 2 years ago
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    Main bhi popat bane ke bad sikha ye concept. So don't worry :D

  20. shivam_bhalla
    • 2 years ago
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    ye concept *sikha

  21. heena
    • 2 years ago
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    janti hu par guys y qn mera ni tha :P some one asked or m physc m apna dimag ni lagati sirf fimrulae se mtlb rakhti hu :P baki to aplog ho na :D :D

  22. heena
    • 2 years ago
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    but one thing surely i m not going to forgot this concept ever :P

  23. shivam_bhalla
    • 2 years ago
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    LOL Remember only the vital Formulas and derice the rest using conservation of 1) Energy 2)linear momentum 3)Angular momentum

  24. shivam_bhalla
    • 2 years ago
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    *derive

  25. heena
    • 2 years ago
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    bhalla u free?? mje isme se ek bhi derivation ni ati frnkly sayin i m so dumb plz help me if u have time tell me how we do derivation :?

  26. shivam_bhalla
    • 2 years ago
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    Sorry not *free exactly but if you want help in any specific derivation, then I can help you

  27. heena
    • 2 years ago
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    actaully in 11 my teacher dosent teach v.well thi topic i mean conservation of laws dats y i m facing problems.. :(

  28. shivam_bhalla
    • 2 years ago
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    Go through the basics of work , gravitational force, kinetic energy, linear momentum, angular momentum, Then for any derivation,you have to apply only these 2 or 3 laws conservation of 1) Energy 2)linear momentum 3)Angular momentum If you have any doubt, you can post a question here and tag me. I will try t help you with that derivation

  29. shivam_bhalla
    • 2 years ago
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    *to

  30. heena
    • 2 years ago
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    ok thnk you guys for being here i have to go to study and yea i ll shivam and one qn @shivam yeterday was ur bday?

  31. shivam_bhalla
    • 2 years ago
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    or @apoorvk

  32. shivam_bhalla
    • 2 years ago
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    @heena , yes

  33. heena
    • 2 years ago
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    dw m yaha sbko tang krne k liye hi hu :D :D and belated bday sorry unable to find u yeterday... many many returns of the belated day... and orry once again for being late to wish u...

  34. shivam_bhalla
    • 2 years ago
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    LOL np :)

  35. heena
    • 2 years ago
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    ok byee...

  36. swalah
    • 2 years ago
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    shiv are u there?

  37. shivam_bhalla
    • 2 years ago
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    yes @swalah

  38. swalah
    • 2 years ago
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    how the angle inpnd?

  39. shivam_bhalla
    • 2 years ago
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    inpnd??

  40. swalah
    • 2 years ago
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    am asking why the angle is independ of the escape velocity?

  41. shivam_bhalla
    • 2 years ago
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    LOL, Ofcourse @swalah , put an angle of projection will increase your escape velocity and escape velocity is the "minimum" velocity required to escape the gravity of earth

  42. shivam_bhalla
    • 2 years ago
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    *putting

  43. swalah
    • 2 years ago
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    @shivam_bhalla putting?

  44. shivam_bhalla
    • 2 years ago
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    LOL, Ofcourse @swalah , "putting " an angle of projection will increase your escape velocity and escape velocity is the "minimum" velocity required to escape the gravity of earth

  45. swalah
    • 2 years ago
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    thanks then what is the angle of that minimum velocity? @shivam_bhalla

  46. shivam_bhalla
    • 2 years ago
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    Ofcourse, 90 degrees

  47. swalah
    • 2 years ago
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    you assume space is spherical yeah?

  48. swalah
    • 2 years ago
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    @shivam_bhalla

  49. shivam_bhalla
    • 2 years ago
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    The gravitational force of the the earth acts spherically as a function of r.

  50. swalah
    • 2 years ago
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    thanks yar bye

  51. shivam_bhalla
    • 2 years ago
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    bye

  52. heena
    • 2 years ago
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    from behalf of @Mani_Jha It doesn't matter whether you throw a bus or a ball into space. Both of them need the same speed 11.6 km/sec to do that(escape velocity of earth). A possible reason of that is that the gravitational field(g) is same for all objects on earth. All objects are in the same field(gravitational) and so require the same velocity to escape the earth. Also, whatever the object be, it has to travel the same distance from the ground to space. Although, it is more difficult to give a bus a speed of 11.6 km/sec than to a ball. Whether you throw it vertically upwards or at an angle, it doesn't matter for the speed. This is because, as shivam_bhalla said, the earth is spherically symmetrical(roughly). Look above at the sky. There is space everywhere at the same distance from you. So, from wherever you throw it, if it has a speed of 11.6 km/sec, it will reach space. , as I think that @shivam_bhalla might have been mistaken.

  53. shivam_bhalla
    • 2 years ago
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    @heena , You are right. :D I was half sleepy . thanks for correcting me. :)

  54. heena
    • 2 years ago
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    no give thanks to @Mani_Jha as i mention in behalf of him i m posting the answer..

  55. heena
    • 2 years ago
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    he is having trouble in posting answer.. :(

  56. shivam_bhalla
    • 2 years ago
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    Thanks @Mani_Jha . and @heena , I was just about to ask about that

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