A satellite is in a circular orbit 21000 km above the Earth’s surface; i.e., it moves on a circular path under the influence of
1 answer:
Answer:
(orbital speed of the satellite) V₀ = 3.818 km
Time (t) = 4.5 × 10⁴s
Explanation:
Given that:
The radius of the Earth is 6.37 × 10⁶ m; &
the acceleration of gravity at the satellite’s altitude is 0.532655 m/s
We can calculate the orbital speed of the satellite by using the formula:
Orbital Speed (V₀) = √(r × g)
radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m
= (2.1 × 10⁷ + 6.37 × 10⁶) m
= 27370000
= 2.737 × 10⁷m
Orbital Speed (V₀) = √(r × g)
Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )
= 3818.215
= 3.818 × 10³
= 3.818 Km
To find the time it takes to complete one orbit around the Earth; we use the formula:
Time (t) = 2 π ×
= 2 × 3.14 ×
= 45019.28
= 4.5 × 10 ⁴ s
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Explanation:
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- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
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