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3 years ago

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If the Hubble Space Telescope is 598 m above the surface of the Earth and is traveling at 7.56 x 103 m/s, how long does it take

the Hubble Space Telescope to orbit the earth? (Radius of the earth RE= 6.38 x 106 m, Mass of the earth ME = 5.98 x 1024 kg)

2 answers:

viva [34]3 years ago

7 0

Answer:

5069.04 seconds

Explanation:

The parameter we are looking for is called the Orbital period of the Hubble Space Telescope.

It is given as:

$T = \sqrt{\frac{4\pi^2r^3 }{GM} }$

where r = radius of orbit of Hubble Space Telescope

G = gravitational constant = $6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}$

M = Mass of earth

We are given that:

r = radius of the earth + distance of HST from earth

r = $6.38 * 10^6 + 598 = 6380598 m$

M = $5.98 * 10^{24} kg$

Therefore, T will be:

$T = \sqrt{\frac{4*\pi^2 * 6380598^3 }{6.67408 * 10^{-11} * 5.98 * 10^{24}}}$

$T = 5069.04 secs$

The orbital period of the Hubble Space Telescope is 5069.04 seconds.

Darya [45]3 years ago

3 0

Answer:

88.25822 Minutes

Explanation:

$T=\frac{S}{V}$ Relates T, time period To S, circumference at height of 598m and V velocity.

S = circumference = $2\pi (h+r)=2\pi (598m+3.38*10^6)=40033930.94meters.$

Substituting all this in our equation gives.

$T = \frac{40033930.94m}{7.56*10^3m/s} =5295.49351s=88.25822minutes.$

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Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

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$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

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T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

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As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

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$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

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