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