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3 years ago

15

A group of college students eager to get to Florida on a spring break drove the 710-mi trip with only minimum stops. They comput

ed their average speed for the trip to be 55.7 mi/h.How many hours did the trip take?

1 answer:

Elis [28]3 years ago

7 0

Answer:

Time taken for trip = 12.74 hour (Approx)

Explanation:

Given:

Distance of trip = 710-mi

Average speed for the trip = 55.7 mi/h

Find:

Time taken for trip = ?

Computation:

⇒ Time = Distance / Speed

⇒ Time taken for trip = Distance of trip / Average speed for the trip

⇒ Time taken for trip = 710-mi / 55.7 mi/h

⇒ Time taken for trip = 12.74 hour (Approx)

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<h3>a)</h3>

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On the other hand, the net force on an object in a centripetal motion should be:

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$m$ is the mass of the object (in this case, that's the mass of the satellite.)
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$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

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$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

<h3>b)</h3>

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Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

<h3>c)</h3>

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$24.6 \times 3600 = 88560\; \rm s$

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$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

<h3>d)</h3>

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$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

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$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

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