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What is the true weight of an object in a geostationary satellite that weighed 10.0N at the north pole?

rooney

edited:

A geostationary satellite is at a height of 36000 KM.

g at that height is given by eq
ma=GmM/R² where R=42000Km

True weight refers to weight at earths surface. So R becomes 1/7th the value (6400Km). So, g increases 49 times.

So, its true weight should be roughly 490N.

iitkgp_bipin

Well done Rohil. Good one.

svj29

for geostationary satellite,
r=42000km
h=36000
mg=10 given

m=g/10

now use the frmula g'=g(R/R+h) whole square......and solve u'll get the crct ansr as per given in HCV.....nw wat remains is jz to put d value....

ramyani

is not svj29 correct ?

rooney

Hes done the same thing. Equation written is wrong.
The 10 N weight given is calculated in the satellite. True weight must be its weight at surface of earth.since its the weight at earth's surface which we refer to as "true weight".
It is (R+h)^2/R^2 in RHS instead of R^2/(R+h)^2.
Weight at earth surface = mg( R+h)^2/R^2 = approx 490 N.

svj29

aaaaaaaaaaaaaarrrrrrrrrghhhhhhhhhhhh.............
cmon jz tell me one dng dat how can g increase when we go up or down............g is max at srface....
rooney as used the wrong formula, the formula has gone upside dwn...
i'm damn sure about my ansr, if u want i cn post a detailed soln of the question again.....

rooney

Dude

Tell me one thing. The weight they have given i.e. 10 N is calculated in the satellite at h=36000 Kms. When you come to the surface , g increases by around 49 times. So, "true weight" is 49 times of 10 N. so, answer is 490 N. and how have i used the wrong formula ? Elaborate.

Ohh and show at one place where i said g increases away from the surface. Read before you say.

svj29

bro u'v misread d questin.......read it again.......its given dat wt of satellite is 10N at northpole........

rooney

Its the definition of "true weight" which I take to be different.

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