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A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours.



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Answer: DSolution: According to kepler 's law T^2 prop R^3(T_1^2)/(T_2^2) = (R_1^3)/(R_2^3) Here R_1 = R+6R = 7RR_2 = 2.5 R+R = 3.5RrArr (24xx24)/(T_2^2) = (7xx7xx7xxR^3)/(3.5xx3.5xx3.5xxR^3) rArrr T_2 = 8.48 hr
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Question Text | A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours. |
Answer Type | Text solution:1 |
Upvotes | 151 |