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beth12345

  • 3 years ago

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

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  1. sgholap100
    • 3 years ago
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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. hosein
    • 3 years ago
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    \[F=GmM/(R _{e}+r)^2\] where M is mass of the Earth & Re is radius of the Earth

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

  4. hosein
    • 3 years ago
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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)

  5. sgholap100
    • 3 years ago
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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

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

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

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

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

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

  11. beth12345
    • 3 years ago
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    ya

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

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

  14. sgholap100
    • 3 years ago
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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

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

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

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

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

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

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

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

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

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

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

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

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

  27. hosein
    • 3 years ago
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    what thing?sgholap100

  28. sgholap100
    • 3 years ago
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    accelaration due to gravity

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

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

  31. sgholap100
    • 3 years ago
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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

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

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

  34. sgholap100
    • 3 years ago
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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

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

  36. sgholap100
    • 3 years ago
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    yes you r right

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

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

  39. sgholap100
    • 3 years ago
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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 ^_^

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