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2 years ago

Determine the distance above Earth's surface to a satellite that completes four orbits per day.

Physics

1 answer:

serg [7]2 years ago

8 0

This question involves the concepts of the time period, orbital radius, and gravitational constant.

The distance of the satellite above the Earth's Surface is "10400 km ".

The theoretical time period of the satellite around the earth can be found using the following formula:

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

where,

T = Time Period of Satellite = $\frac{1}{frequency} = \frac{1}{4\ orbits/day}\frac{3600*24\ s}{1\ day} = 21600\ s$

R = Orbital Radius = ?

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

Therefore,

$\frac{(21600\ s)^2}{R^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\R = \sqrt[3]{\frac{4.66\ x\ 10^8\ s^2}{9.91\ x\ 10^{-14}\ s^2/m^3}} \\R = 1.675\ x\ 10^7\ m = 1.68\ x\ 10^4\ km$

Now, this orbital radius is the sum of the radius of the earth (r) and the distance of satellite above earth's (h) surface:

R = r + h

1.68 x 10⁴ km = 0.64 x 10⁴ km + h

h = 1.68 x 10⁴ km - 0.64 x 10⁴ km

Learn more about the orbital time period here:

brainly.com/question/14494804?referrer=searchResults

The attached picture shows the derivation of the formula for orbital speed.

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2 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

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3 years ago

During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu

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Answer:

The horizontal component of the velocity is 188 m/s

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Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

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195 m/s · cos 15° = vx

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The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

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The vertical component of the velocity is 50 m/s.

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3 years ago

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