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allsm [11]
3 years ago
6

A satellite moves in a circular orbit at a constant speed v0 around Earth at a distance R from its center. The force exerted on

the satellite is F0. An identical satellite orbits Earth at a distance of 3R from the center of the Earth. How does the tangential speed, vT, of the satellite at distance 3R compare to the speed v0 of the satellite at R
Physics
1 answer:
postnew [5]3 years ago
6 0

Answer:

vT = v0/3

Explanation:

The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R

Since F = F0 = F'

GMm/R² = m(v0)²/R

GM = (v0)²R  (1)

Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R

Since F1 = F'

GMm/27R² = m(vT)²/3R

GM = 27(vT)²R/3

GM = 9(vT)²R (2)

Equating (1) and (2),

(v0)²R = 9(vT)²R

dividing through by R, we have

9(vT)² = (v0)²

dividing through by 9, we have

(vT)² = (v0)²/9

taking square-root of both sides,

vT = v0/3

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A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14
Anestetic [448]

Answer:

3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

\omega_f^2 - \omega_i ^2 =2 \alpha \theta

where

\omega_f = 3.14\cdot 10^4 rad/s is the final angular speed

\omega_i = 1.10 \cdot 10^4 rad/s is the initial angular speed

\theta= 2.00 \cdot 10^4 rad is the angular distance covered

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Re-arranging the formula, we can find \alpha:

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2

Now we want to know the time the bit takes starting from rest to reach a speed of \omega_f=7.85\cdot 10^4 rad/s. So, we can use the following equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where:

\alpha=2.16\cdot 10^4 rad/s^2 is the angular acceleration

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\omega_i = 0 is the initial speed

t is the time

Re-arranging the equation, we can find the time:

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{7.85\cdot 10^4 rad/s-0}{2.16\cdot 10^4 rad/s^2}=3.63 s

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3 years ago
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