Orbital Radius of Earth Satellite: Solve for r

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To find the orbital radius of an Earth satellite with a period of 1 day, the formula r^3 = (GMT^2) / (4(pi)^2) is used. The gravitational constant (G) is 6.67 x 10^(-11) m^3 kg^(-1) s^(-2) and the mass of Earth (M) is approximately 6 x 10^(24) kg. The error in the calculation arises from not converting the period (T) from days to seconds, which is crucial for accuracy. After correcting for this, the expected orbital radius aligns with the book's value of 42.2 Mm. Proper unit conversion is essential for accurate results in orbital mechanics.
Bernie Hunt
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Find the orbit radius of an Earth satellite in a circular orbit if the period if 1d.

I'm at a dead end. So far I have;

r^3 = (GMT^2) / (4(pi)^2)

r^3 = (6.67*10^(-11) * 6*10^(24)) / (4 * pi^2)

and I get

r = 21642

The book says 42.2Mm. I'm only off by a couple of million meters, hahaha.

Where am I going wrong?

Thanks,
Bernie
 
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You need to use the value of T in seconds for the constants you used to be correct.
 
Argh! Thud, thud, thud ...

That's the sound of me banging my head against the wall!

Thanks for the quick reply.

Bernie
 
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