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2 months ago

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In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P

luto already was known to have a large satellite, Charon, orbiting at 19,600 km with an orbital period of 6.39 days.
Required:
Assuming that the satellites do not affect each other, find the orbital periods T1 and T2 of the two small satellites without using the mass of Pluto.

1 answer:

inna [3.1K]2 months ago

8 0

Answer:

The period for the first satellites is 24.46 days, while the second satellites have a period of 37.67 days

Explanation:

Provided:

Distance of the first satellites $r_{sat1} = 48000 \times 10^{3}$ m

Distance of the second satellites $r _{sat2} = 64000 \times 10^{3}$ m

Distance of Charon $r_{c} = 19600 \times 10^{3}$ m

Orbit period of Charon $T_{c} = 6.39$ days

According to Kepler's third law,

the square of the orbital period correlates to the cube of the semi-major axis.

$T^{2} = r^{3}$

$\frac{T}{r^{\frac{3}{2} } } = constant$

For the first satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }$

${T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat1} = 24.46$ days

For the second satellites,

$\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }$

${T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}$

$T_{sat2} = 37.67$ days

Thus, the orbital period for the first satellites is 24.46 days and for the second satellites, it is 37.67 days

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