Question

Sonda okrąża Ziemie w czasie t=90 min na wysokości h=430km.Promień Ziemi ma wartość R=6370 km . Oblicz przyspieszenie dośrodkowe sondy . Proszę o pomóżcie :D

Answer 1

T = 90 min = 90 *60 s = 5400 s = 54 *10² s

h = 430 km

R = 6370 km

 R₁ = R + h = 6800 km = 6900000 m = 68 *10⁵ m

a = ?

 

a = V² / R ₁

a = V²/R+h

V = 2πR₁ / T

V = 2π (R + h ) / T

V² = 4π² (R + h )² / T²

 

a = 4π² (R + h )² / T² /R + h

a = 4π²(R + h ) ( R + h ) / T² * 1 / (R+h)

a = 4 π² ( R + h ) /T²

 

a = 4 * (3,14)² * 68 *10⁵ m /( 54*10² s )²

a = 4 *9,86 * 68 *10⁵ m /2916 *10⁴ s²

a ≈ 9,2 m / s²

Answer 2

dane:

T = 90 min = 90 x 60s = 5 400 s = 5,4 x 10^3 s

h = 430 km

R = 6370 km

r = h+R = 430km+6370km = 6800 km = 6 800 000 m = 6,8 x 10^6

szukane:

ar = ?

 

ar = v^2 /r 

       v = 2TTr/T

       v^2 = 4TT^2 /T2

ar = 4TT^2 *r^2 /T^2 : r = 4TT^2 *r/T^2 = 4 x 3,14^2 x 6,8 x 10^6/(5,4 x 10^3 s)^2 =

    = 39,4 x 6,8/29,16 m/s2 = 9,187 m/s2

ar = 9,2 m/s2

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