Question

given that Earth's surface gravitational field strength has a magnitude of 9.8N/kg, determine the distance above Earth's surface at which the magnitude of the field strength is 3.2N/kg

Answer 1

\[F=Gm _{1}m _{2}/r ^{2}\]
use this to see that the force changes invers proportionaly to square of distance

Answer 2

m1 is mass of earth?
G is gravitational constant?
what is m2?

Answer 3

m2 would be one kg, but you don't need that.
9,8/r2 = 3.2

Answer 4

the answer is 0.75.

Answer 5

sqroot(9.8/3.2)=1.75

Answer 6

i would say 1,75 earth radius

Answer 7

M= earth mass
m= 1 kg
G = universal constant
r = distance
\[9,8 = GMm/r _{1}^{2}\]
\[3.2 = GMm/r _{2}^{2}\]
deviding bouth:
\[9,8/3,2 = r _{2}^{2}/ r _{1}^{1}\rightarrow 1,75 r _{1} =r _{2}\]

Answer 8

ok thanks

Answer 9

i got it

Answer 10

give medal ?