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3 years ago

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]3 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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$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6$

E₀ is a constant and each surface is equal to each other, so: $S_1=S_2=S_i=S$

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$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c$

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