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nadya68 [22]
3 years ago
14

As planets with a wide variety of properties are being discovered outside our solar system, astrobiologists are considering whet

her and how life could evolve on planets that might be very different from earth.One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun.This planet has been found to have 2.3 times the earth's diameter and 7.9 times the earth's mass.For planets in this size range, computer models indicate a relationship between the planet's density and composition. Observations of this planet over time show that it is in a nearly circular orbit around its star and completes one orbit in only 9.5 days. How many times the orbital radius r of the earth around our sun is this exoplanet's orbital radius around its sun? Assume that the earth is also in a nearly circular orbit
Physics
1 answer:
ra1l [238]3 years ago
3 0

Answer:

R₁  = 14.44 R₂

Explanation:

For sun-earth system

GMm/R² = mω²R

GM =ω²R³

GM₁ =ω₁²R₁³

For star - planet system

GM₂ =ω₂²R₂³

Dividing

M₁ / M₂ = ω₁²R₁³ / ω₂²R₂³

R₁³ /R₂³ = M₁ω₂² / ω₁²M₂

= M₁T₁²/T₂²M₂

Given

M₁ / M₂ = 1 / 0.7

T₁ = 365 days

T₂ = 9.5 days

R₁³ /R₂³ = M₁²365² /9.5² (.7M₁ )²

(R₁ /R₂)³ =365² / 9.5² X .7²

= 3012

R₁ /R₂

= 14.44

R₁  = 14.44  R₂

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Explanation:

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Newton's first law states that objects do not change their velocity unless acted upon
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In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)
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u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

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Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm

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7 0
3 years ago
A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear
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Answer:

70.07 Hz

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v_s=27.76 m/s

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f_s=70.07 Hz

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3 years ago
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