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3 years ago

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The Hubble Space Telescope in orbit above the Earth has a 2.4 m circular aperture. The telescope has equipment for detecting ult

raviolet light. What is the minimum angular separation between two objects that the Hubble Space Telescope can resolve in ultraviolet light of wavelength 95 nm?

1 answer:

kupik [55]3 years ago

7 0

Answer:

minimum angular separation between two objects is; θ = 4.829 x 10^(-8) rad

Explanation:

The minimum angle for resolving two adjacent point objects is proportional to the wavelength of the light and is inversely proportional to the diameter of the aperture and it's given by the formula;

θ = 1.22λ/D

Where;

θ is the resolved angle

λ is the wavelength

D is the diameter of the optics

We are given;

λ = 95nm = 95 x 10^(-9)m

D = 2.4m

Thus ;

θ = (1.22(95 x 10^(-9))/2.4

θ = 4.829 rad

Thus, minimum angular separation between two objects is; θ = 4.829 x 10^(-8) rad

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It is the energy due to the position.

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Explanation:

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$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

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It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

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$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

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$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

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Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

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$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

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