Question

find the weight of an object at a height of 6,400 km above the earth's surface. the weight of the object at the surface of the earth is 20N and the radius of the earth is 6,400 km.
today is my science exam please solve

Answer 1

g(h) = g{r/r+h}^2
g(h) = 9.8 . {6400/6400+6400}^2
g(h) = 9.8 . (0.5)^2
g(h) = 9.8 × 0.25
g(h) = 2.45

weight of the body equals to Mg....
mass = 20N/9.8 = 2.04kg
g(h) equals to 2.45
weight equals to 2.04×2.45 = 5N

Answer 2

Hey hi friend!!!
Well this question is related to gravity for this question u would need this equation
F=GMm/r^2
here F=gravitational force ,
G= universal gravitational constant ,
M is mass of the object 1 (here earth)
and m is mass of object 2( here the given object )
and r is the distance between the centres of the two objects

and here i am going to show u a short cut method
We know that GMm is never going to change like G is universal constant and M and m is always a constant in this case so let us assume that
GMm=k (which is a constant)
So in first case
we are given the weight of the object i.e. 20N and note that this weight is the gravitational force acting on the object
therefore F=20N
So the distance given is 6400 i.e r = 6,400km
hence substituting the values we get
20=k/(6400)^2
k=20*6400*6400
k= 8192*10^5 ____________(1)


so we know value of k
so let us calculate the weight when it is 6400 km above the earth's surface
in that case r=6400+6400=12800km because here we are 6400 km high above the earth's surface and earths surface is itself 6400km away from its center
so r = 12800km
so by using the formula
F=k/r^2
F=8192*10^5/(12800)^2
F=8192*10^5/(12800*12800)
F=8192*10^5/(16384*10^4)
F=0.5*10
F=5N
Hence the weight of the object
at a height of 6,400 km above the earth's surface is 5 N

Hope u got it :-)
And Best Of Luck for your exam