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Question Number 104143 by aurpeyz last updated on 19/Jul/20

A Satellite orbits the esrth in a circle of rsdius 8000km. At that distance from the earth g=6.2m/s2. The velocity of the satelliege is?

$$\mathrm{A}\:\mathrm{Satellite}\:\mathrm{orbits}\:\mathrm{the}\:\mathrm{esrth}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{rsdius}\:\mathrm{8000km}.\:\mathrm{At}\:\mathrm{that}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{earth}\:\mathrm{g}=\mathrm{6}.\mathrm{2m}/\mathrm{s2}.\:\mathrm{The}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{satelliege}\:\mathrm{is}? \\ $$

Commented by aurpeyz last updated on 19/Jul/20

pls help

$$\mathrm{pls}\:\mathrm{help} \\ $$

Answered by OlafThorendsen last updated on 19/Jul/20

F^→  = mg^→  = ma_N ^→ = m(v^2 /R)u^→   ⇒ v = (√(gR)) = (√(6,2×8.10^6 ))  v = 7,042 km/s

$$\overset{\rightarrow} {\mathrm{F}}\:=\:{m}\overset{\rightarrow} {{g}}\:=\:{m}\overset{\rightarrow} {{a}}_{\mathrm{N}} =\:{m}\frac{{v}^{\mathrm{2}} }{\mathrm{R}}\overset{\rightarrow} {{u}} \\ $$$$\Rightarrow\:{v}\:=\:\sqrt{{g}\mathrm{R}}\:=\:\sqrt{\mathrm{6},\mathrm{2}×\mathrm{8}.\mathrm{10}^{\mathrm{6}} } \\ $$$${v}\:=\:\mathrm{7},\mathrm{042}\:{km}/{s} \\ $$

Commented by aurpeyz last updated on 20/Jul/20

thnks

$$\mathrm{thnks} \\ $$

Commented by aurpeyz last updated on 01/Aug/20

pls how did you get R as 8.10^6 ?

$${pls}\:{how}\:{did}\:{you}\:{get}\:{R}\:{as}\:\mathrm{8}.\mathrm{10}^{\mathrm{6}} ? \\ $$

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