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hoa [83]
2 years ago
10

A satellite orbits the Earth (mass = 5.98 x 1024 kg) once every = 43200 s. At what radius does the satellite orbit?

Physics
1 answer:
kap26 [50]2 years ago
7 0

Answer:

26621 km

Explanation:

We are given;

Mass: m = 5.98 x 10^(24) kg

Period; T = 43200 s

Formula for The velocity(v) of the satellite is:

v = 2πR/T

Where R is the radius

Formula for centripetal acceleration is;

a_c = v²/R

Thus; a_c = (2πR/T)²/R = 4π²R/T²

Formula for gravitational acceleration is:

a_g = Gm/R²

Where G is gravitational constant = 6.674 × 10^(-11) m³/kg.s²

Now the centripetal acceleration of the satellite is caused by its gravitational acceleration. Thus;

Centripetal acceleration = gravitational acceleration.

Thus;

4π²R/T² = Gm/R²

Making R the subject gives;

R = ∛(GmT²/4π²)

Plugging in the relevant values;

R = ∛((6.674 × 10^(-11) × 5.98 x 10^(24) × 43200²)/(4 × π²))

R = 26.621 × 10^(6) m

Converting to km, we have;

R = 26621 km

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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
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Therefore the ratio of diameter of the copper to that of the tungsten is

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The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

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R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

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