Question

A 620 kg satelite above Earth's surface experiences a gravitational field strenght at magnitude 4.5 N/Kg
a)knowing earth's rad ,how far above earths surface is the satelite? (use ratio and proportion) ans: 3.0 times 10^6 I dont get the answer however it looks easy

Answer 1

Height h=√[GmM/F] -R

Answer 2

Whats is F?

Answer 3

So you know that the gravitational force is dictated by Newtons Law of gravitation (equation a). You also know that the height they are asking you for is (hans) the height from the center of the earth (h) - the radius of the earth (rearth)(equation b). Finally you know that the gravitational force is also the gravitational field (G.F.) times the mass of the satellite (equation c). If you substitu equation b and c into equation a and you solve for hans, then you will get the right answer. Below are the maths.





















































Equation (a)
\[F _{c}=\frac{ m _{1}m _{earth} }{ h ^{2} }G\]Equation (b)\[h _{ans}=h-r _{earth}\]Equation (c)
\[F _{c}=G.F.m _{1}\]Final equation:\[G.F.m _{1}=\frac{ m _{1}m _{earth} }{ (h _{ans}+r _{earth}) }G \rightarrow h _{ans}=\sqrt{\frac{ m _{earth} }{ G.F. }G}-r _{earth}\]Solution:\[h _{ans}=\sqrt{\frac{ 5.97 \times 10^{24}kg }{ 4.5\frac{ N }{ kg } }6.67\times 10^{-11}\frac{ Nm ^{2} }{ kg ^{2} }}-6371000m=3 \times 10^{6}m\]

Answer 4

Thank you very much