Zooming around the Sun!
If we would build a satellite sufficiently heat-resistant to orbit the sun directly above its surface, in how much time would that satellite complete a single orbit?
Give your answer in hours, rounded to the nearest integer.
Data to be used:
sun’s radius earth’s distance to sun = 2 1 5 , one hour one year = 8 7 6 6 .
The answer is 3.
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2 solutions
The question isn't really answerable, since no stats are given in regards to how fast the satellite would be orbiting. You can't just assume it'll be at the same rate as the earth.
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In order to remain in orbit around the sun, it has to be orbiting at just the right speed (and a whole lot faster than the earth!). If not, it would fall into the sun or travel in an elliptical orbit rather than staying close to the surface.
The assumption I make is that the satellite is orbiting the same central mass as the earth (namely, the sun). As a result, the gravitational acceleration of earth and satellite are connected as a ⊕ a s a t = r s a t 2 r ⊕ 2 = 2 1 5 2 ; this follows from Newton's universal law of gravitation.
To maintain a circular orbit, the speed must obey the centripetal acceleration equation, a = r v 2 ∴ v = a r . Thus v ⊕ v s a t = a ⊕ a s a t ⋅ r ⊕ r s a t = r s a t 2 r ⊕ 2 ⋅ r ⊕ r s a t = r s a t r ⊕ = 2 1 5 . Thus the satellite must be traveling 2 1 5 ≈ 1 4 . 5 times faster than the earth.
Finally, the orbital period is T = 2 π r / v , so that T ⊕ T s a t = r ⊕ r s a t ⋅ v s a t v ⊕ = r ⊕ r s a t ⋅ r ⊕ r s a t = ( r ⊕ r s a t ) 3 / 2 = 2 1 5 − 3 / 2 . This means that the satellite's orbital period is 0.000317 times that of the earth.
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Making 8766 as time period of earth was smart . i.e. average time would be equal to [24x365x0.75 + 24x366x0.25] which is 8766. Nice use of weighted mean !
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@Mayank Hooda – I typically work with 1 year = 365.25 days...
One of the most easiest maths I have ever done. It's all about the Keplarian motion which states T²=r³ ( T for time and r for radius). So, T²=(1/215)³ T=2.78 hours or 3 hours.
Of course, this problem is not about the math... it is about getting the concepts right.
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I, for one, loved this problem. Your problems are endlessly entertaining, Arjen.
Since the earth and the satellite orbit the same central mass (the sun), we can use Kepler's Third Law:
r ⊕ 3 r s a t 3 = T ⊕ 2 T s a t 2 ; T s a t = T ⊕ ⋅ ( r ⊕ r s a t ) 3 / 2 ; T s a t = 8 7 6 6 hours ⋅ ( 2 1 5 1 ) 3 / 2 ; T s a t = 2 . 7 8 ≈ 3 hours .