beth12345
  • beth12345
What would be the gravitational field strength 2000km above the Earth? 50km above the Earth?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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.
anonymous
  • anonymous
\[F=GmM/(R _{e}+r)^2\] where M is mass of the Earth & Re is radius of the Earth
beth12345
  • beth12345
what is Gm and r?

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anonymous
  • anonymous
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)
anonymous
  • anonymous
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
anonymous
  • anonymous
i think hosein it is h because the question stats that is about the altitude
beth12345
  • beth12345
uhm, I don't really know the numbers to plug into the equation
anonymous
  • anonymous
yea here r is h
anonymous
  • anonymous
you can't solve this,bath?
anonymous
  • anonymous
so u can put 2000km as h or 5o as h but u cannot solve it
beth12345
  • beth12345
ya
beth12345
  • beth12345
i don't know what m is
anonymous
  • anonymous
m is mass of 2nd body
anonymous
  • anonymous
m is the mass of a any object on the earth surface or the object above the earth surface e.g a satellite
beth12345
  • beth12345
do you know what value I would plug into it?
anonymous
  • anonymous
mass of an object e.g 40 kg but is negligible as compared to earths mass
anonymous
  • anonymous
G=6.67*10(-11) N(m/kg)^2 M = 5.9722 × 1024 kg
anonymous
  • anonymous
subtitute the values given if u have any word problem
anonymous
  • anonymous
you ought be consider that in formula
beth12345
  • beth12345
ok so would the equation then become \[G= 6.67\times10^{-11}\] \[M= 5.98 \times10^{24}\] \[R e = 6.38 \times10^{6 }\] \[h= (2000)or (50)\]
anonymous
  • anonymous
yea put it in formula
anonymous
  • anonymous
you should these standard values in the formula
beth12345
  • beth12345
and then \[((6.67\times10^{-11}) \times 40) \div ((6.38 \times 10^{6}) + 2000)\]
anonymous
  • anonymous
where is Earth mass? and denominator has power 2
anonymous
  • anonymous
but accelaration due to gravity too comes in the formula
anonymous
  • anonymous
we forgot the main thing
anonymous
  • anonymous
what thing?sgholap100
anonymous
  • anonymous
accelaration due to gravity
anonymous
  • anonymous
ican't understand your goal?can you explane
anonymous
  • anonymous
you can use gRe^2 instead of GMe
anonymous
  • anonymous
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
anonymous
  • anonymous
your goal is?\[mg=GMm/R _{e}^2\] am i right?
beth12345
  • beth12345
\[((6.67 \times 10 ^{-11}) \times 40) \times 5.98 \times 10 ^{24}) \div(6.38 \times 10 ^{6} +2000)^{2} \]
anonymous
  • anonymous
oh yeah i am sorry its was my mistake i was considering Effect of altitude on gravity @hosein u r right
anonymous
  • anonymous
ithink it's right.
anonymous
  • anonymous
yes you r right
beth12345
  • beth12345
so would the answer be 391.7 ?
beth12345
  • beth12345
ok ya I think I got it, thanks guys :)
anonymous
  • anonymous
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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