Question

An artificial satellite is in a circular orbit d= 740.0 km above the surface of a planet of radius r = 4.25 x 108 km. The period of revolution of the satellite around the planet is T = 1.15 hours. What is the average density of the planet? density = kg/m3

Answer 1

GmM/(r+h)^2 = m*w^2* (r+h)

w^2 = GM /(r+h)³

T = 2pi /w  

so

T =2pi sqrt ( r+h)³ /GM)

T = 1.15 hour = 1.15*60*60 =4140 sec

4140 = 2*pi * sqrt( (4.25*10^6+740*10^3)^3 /(6.67*10^-11*M))

(4.25 106 + 740_103) 4140 = 2 6 .67 10-11 M Enlarge Customize A Plain Text Result: 4140 = 8.57566 x 1015

from here M = 4.29*10²⁴ kg

mass = density *volume

densty = mass /volume = 4.29*10²⁴ / ( 4/3*pi*(4.25*10^6)^3)

Input interpretation: 4.29 1024 (4.25106) Result: 13341.41...

13341.41 kg/ m^3 answer

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