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

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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.
\[F=GmM/(R _{e}+r)^2\] where M is mass of the Earth & Re is radius of the Earth
what is Gm and r?

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Other answers:

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