a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

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

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ 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 $v$ :

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

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

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

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

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

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

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

Convert $24.6\; \rm \text{hours}$ to seconds:

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

Solve the equation for $R$ :

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

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

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

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

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

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

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

5 0

4 years ago

A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po

Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + $0.5mV1^{2}$ = mgh2 + $0.5mV2^{2}$

where

speed at highest point = V2

speed at lowest point = V1

mass of the girl and swing = m

at the highest point, the speed is minimum (V1 = 0)

at the lowest point the speed is maximum (V2 is the maximum speed)

therefore the equation becomes mgh1 + $0.5mV1^{2}$ = mgh2

m(gh1 + $0.5V1^{2}$ ) = m(gh2)

gh1 + $0.5V1^{2}$ = gh2

V1 = $\sqrt{\frac{gh2 - gh1}{0.5}}$

now we can substitute all required values into the equation above.

V1 = $\sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}$

V1 = $\sqrt{\frac{32.34}{0.5}}$

V1 =8.1 m/s

8 0

3 years ago

As a transverse wave travels through a rope from left to right, the parts of the rope _______.

kupik [55]

<h3><u>Answer;</u></h3>

Are moving up and down.

As a transverse wave travels through a rope from left to right, the parts of the rope <u>are moving up and down</u>.

<h3><u>Explanation;</u></h3>

Transverse waves occur when a disturbance causes oscillations perpendicular to the propagation, that is the direction of energy transfer.
<em><u>Particles of the medium move perpendicular to the direction the transverse wave itself is moving. For example, if the wave is moving to the right, the particles of the medium are moving up and down.</u></em>
<em><u>Therefore, as a transverse wave travels through a rope from left to right, the parts of the rope are moving up and down.</u></em>

8 0

4 years ago

A fish finder uses a sonar device that sends 20,000-Hz sound pulses downward from the bottom of the boat, and then detects echoe

mrs_skeptik [129]

Answer: The minimum time between pulses (in fresh water)

= 0.3106 s

Explanation:

To calculate the speed of echoes sounds, we will use

Speed = 2x/t

Where x = distance and t = total time.

Please note that:

sound travels at 343 m/s in air. But it travels at 1,481 m/s in water (almost 4.3 times as fast as in air);

Total distance = 2 × 230 = 460m

Using speed of sound in water

1481 = 460/t

t = 460/1481

t = 0.3106

3 0

3 years ago