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tekilochka [14]
4 years ago
8

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

om Saturn to the sun and then calculate it.
Physics
2 answers:
d1i1m1o1n [39]4 years ago
3 0
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
kari74 [83]4 years ago
3 0

Answer:

For astronomical objects, the time period can be calculated using:

T² = (4π²a³)/GM

where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)

Thus,

T² = a³

a = ∛(29.46²)

a = 0.67 AU

1 AU = 1.496 × 10⁸ Km

0.67 * 1.496 × 10⁸ Km

= 1.43 × 10⁹ Km

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<h3>In vector form;</h3>

The initial velocity is:

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The acceleration is:

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Using the first equation of motion:

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Then, we can replace their values into the equation of motion in order to determine the speed:

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+4 \times 0.5 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big(0.7 ( -cos 25^0 \hat x + sin 25^0 \hat y )+2.0 ( cos 41^0 \hat x - sin 41^0 \hat y )\Big)}

\mathbf{v^{\to} =\Big( ( -0.7 cos 25^0 \hat x + 0.7 sin 25^0 \hat y )+( 2.0cos 41^0 \hat x - 2.0sin 41^0 \hat y )\Big)}

Collect like terms:

\mathbf{v^{\to} =\Big( (2.0cos 41^0 -0.7 cos 25^0   )\hat x+(  0.7 sin 25^0 - 2.0sin 41^0 )\Big)\hat y}

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\mathbf{v^{\to} =\sqrt{(0.87500 )^2 +( 1.01629 )^2}}

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\mathbf{v^{\to} =\sqrt{1.79848}}

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Learn more about vectors here:

brainly.com/question/17108011?referrer=searchResults

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