How did the textbook get this answer? (satellite motion question)

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The discussion centers on solving a satellite motion problem using the formula T = 2πr/v. The initial calculations for velocity and radius are presented, with a suggestion to use the Earth's radius for the satellite's orbit radius in part B. Participants agree that applying the result from part A, specifically v = √(GM/r), into the equation for T is the correct approach. This method allows for solving for the radius and subsequently the velocity. The conversation highlights the importance of consistent radius usage in satellite motion calculations.
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Homework Statement
Suppose we want to place a weather satellite into a circular orbit 300 km above the earth's surface. The earth's radius is 6380 km= 6.38E6 m, and its mass is 5.98E24 kg
A. What speed, period, and radial acceleration must it have?
B. Find the speed and orbit radius for an earth satellite that has a period of 1 day (86400 s).
Relevant Equations
(all answers are rounded) I understand how to get part A as
v= sqrt((6.67E-11*5.98E24)/(3.0E5+6.38E6))= 7730 m/s
T= 2π(6.68E6)/7730= 5430 s
a= (7730^2)/6.68E6= 8.95 m/s^2
I don't know what I'm doing wrong, but I'm not getting the same answer as the textbook for part B (v= 3.07E3 m/s; r= 4.23E7 m). Can someone please explain how to do part B?
Below is my attempt at the problem:
T= 86400= 2πr/v
v= 2π(6.68E6)/86400= 4.86E7 m/s
r= (86400)(7730)/2π= 1.06E8 m
 
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It seems to me that you are using the Earth's radius as the satellite's orbit radius for part B. You are correct in your approach to part A. I suggest doing the same in part B.

take your result from part A (namely ##v = \sqrt{\frac{GM}{r}}##) and plug it into ##T = \frac{2 \pi r}{v}## to get

##T = 2 \pi r \sqrt{\frac{r}{GM}}## solve for ##r## and then solving for ##v## is straight forward.
 
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