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beth12345

What would be the gravitational field strength 2000km above the Earth? 50km above the Earth?

  • 2 years ago
  • 2 years ago

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  1. sgholap100
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    Gravity decreases with altitude, since greater altitude means greater distance from the Earth's center. so the gravity at 2000 km above the earth surface will be very less as compare to 50 km above the earth surface.

    • 2 years ago
  2. hosein
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    \[F=GmM/(R _{e}+r)^2\] where M is mass of the Earth & Re is radius of the Earth

    • 2 years ago
  3. beth12345
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    what is Gm and r?

    • 2 years ago
  4. hosein
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    r is distance of mass to Earth and G is gravitation constant that equal to 6.63*10^34(i'm not sure about G's value but i think that.many of physics book has this eg holliday)

    • 2 years ago
  5. sgholap100
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    G is the gravitational constant and m is the mass of any object on the earth surface it is not r it is h which means

    • 2 years ago
  6. sgholap100
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    i think hosein it is h because the question stats that is about the altitude

    • 2 years ago
  7. beth12345
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    uhm, I don't really know the numbers to plug into the equation

    • 2 years ago
  8. hosein
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    yea here r is h

    • 2 years ago
  9. hosein
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    you can't solve this,bath?

    • 2 years ago
  10. sgholap100
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    so u can put 2000km as h or 5o as h but u cannot solve it

    • 2 years ago
  11. beth12345
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    ya

    • 2 years ago
  12. beth12345
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    i don't know what m is

    • 2 years ago
  13. hosein
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    m is mass of 2nd body

    • 2 years ago
  14. sgholap100
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    m is the mass of a any object on the earth surface or the object above the earth surface e.g a satellite

    • 2 years ago
  15. beth12345
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    do you know what value I would plug into it?

    • 2 years ago
  16. sgholap100
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    mass of an object e.g 40 kg but is negligible as compared to earths mass

    • 2 years ago
  17. hosein
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    G=6.67*10(-11) N(m/kg)^2 M = 5.9722 × 1024 kg

    • 2 years ago
  18. sgholap100
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    subtitute the values given if u have any word problem

    • 2 years ago
  19. hosein
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    you ought be consider that in formula

    • 2 years ago
  20. beth12345
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    ok so would the equation then become \[G= 6.67\times10^{-11}\] \[M= 5.98 \times10^{24}\] \[R e = 6.38 \times10^{6 }\] \[h= (2000)or (50)\]

    • 2 years ago
  21. hosein
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    yea put it in formula

    • 2 years ago
  22. sgholap100
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    you should these standard values in the formula

    • 2 years ago
  23. beth12345
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    and then \[((6.67\times10^{-11}) \times 40) \div ((6.38 \times 10^{6}) + 2000)\]

    • 2 years ago
  24. hosein
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    where is Earth mass? and denominator has power 2

    • 2 years ago
  25. sgholap100
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    but accelaration due to gravity too comes in the formula

    • 2 years ago
  26. sgholap100
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    we forgot the main thing

    • 2 years ago
  27. hosein
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    what thing?sgholap100

    • 2 years ago
  28. sgholap100
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    accelaration due to gravity

    • 2 years ago
  29. hosein
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    ican't understand your goal?can you explane

    • 2 years ago
  30. hosein
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    you can use gRe^2 instead of GMe

    • 2 years ago
  31. sgholap100
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    as the question say that the object is above earth surface accelaration due to gravity(g=9.8m/s^2) too comes then the formula becomes

    • 2 years ago
  32. hosein
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    your goal is?\[mg=GMm/R _{e}^2\] am i right?

    • 2 years ago
  33. beth12345
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    \[((6.67 \times 10 ^{-11}) \times 40) \times 5.98 \times 10 ^{24}) \div(6.38 \times 10 ^{6} +2000)^{2} \]

    • 2 years ago
  34. sgholap100
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    oh yeah i am sorry its was my mistake i was considering Effect of altitude on gravity @hosein u r right

    • 2 years ago
  35. hosein
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    ithink it's right.

    • 2 years ago
  36. sgholap100
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    yes you r right

    • 2 years ago
  37. beth12345
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    so would the answer be 391.7 ?

    • 2 years ago
  38. beth12345
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    ok ya I think I got it, thanks guys :)

    • 2 years ago
  39. sgholap100
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    welcome but do one thing neglect the m and h in the formula as they are very small as compared to M and R OK ^_^

    • 2 years ago
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