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A 10 kg satellite circles earth once every 2 hours in an orbit having a radius of 8000 km . Assuming that Bohr's angular momentum postulate applies to a satellite just as it does to an electron in the hydrogen atom, then the quantum number of the orbit of satellite is .?
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A 10 kg satellite circles earth once every 2 hours in an orbit having ...
Introduction:
In order to determine the quantum number of the orbit of the satellite, we can apply Bohr's angular momentum postulate, which states that the angular momentum of an electron in an atom is quantized and can only take on certain discrete values. We can extend this concept to the satellite orbiting the Earth and use the given information to find the quantum number associated with its orbit.

Given information:
Mass of the satellite (m) = 10 kg
Orbit radius (r) = 8000 km = 8,000,000 meters
Time period of orbit (T) = 2 hours = 7200 seconds

Calculating angular momentum:
The angular momentum (L) of a satellite in orbit can be calculated using the formula:
L = mvr
where m is the mass of the satellite, v is its linear velocity, and r is the orbit radius.

The linear velocity of the satellite can be found using the formula:
v = 2πr/T
where T is the time period of the orbit.

Substituting the values into the formulas, we get:
v = 2π(8,000,000)/(7200) ≈ 2790.89 m/s
L = (10)(2790.89)(8,000,000) ≈ 2.2327128 x 10^14 kg·m²/s

Applying Bohr's postulate:
According to Bohr's angular momentum postulate, the angular momentum of an electron in an atom is quantized and can only take on certain discrete values, given by the formula:
L = nħ
where n is the quantum number and ħ (h-bar) is the reduced Planck's constant.

We can rearrange this formula to solve for the quantum number:
n = L/ħ

Calculating the quantum number:
The reduced Planck's constant (ħ) is given by:
ħ = h/(2π)
where h is the Planck's constant.

Substituting the value of Planck's constant (h = 6.626 x 10^-34 J·s) into the formula, we get:
ħ = (6.626 x 10^-34)/(2π) ≈ 1.0545718 x 10^-34 J·s

Now, we can substitute the values of L and ħ into the formula for the quantum number:
n = (2.2327128 x 10^14)/(1.0545718 x 10^-34) ≈ 2.117146 x 10^48

Conclusion:
The quantum number of the orbit of the satellite is approximately 2.117146 x 10^48. This demonstrates that Bohr's angular momentum postulate can be applied to the satellite just as it applies to an electron in the hydrogen atom, assuming the satellite's orbit follows classical mechanics.
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A 10 kg satellite circles earth once every 2 hours in an orbit having a radius of 8000 km . Assuming that Bohr's angular momentum postulate applies to a satellite just as it does to an electron in the hydrogen atom, then the quantum number of the orbit of satellite is .?
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