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Gre4nikov
1 month ago
5

If the satellite was placed in an orbit three times farther away, about how long would it take to orbit the Earth once? Answer i

n days, rounding to one significant figure.
Physics
2 answers:
kicyunya [3.2K]1 month ago
7 0

Since satellites are often classified based on their altitude, I will focus on the lowest type, called LEO, which orbits at an altitude of 320-800 km above the surface of the Earth.

Satellites in LEO travel at a speed of 27,359 km/h to complete an orbit around the Earth in 90 minutes. LEO also experiences a delay time of 10 ms (the duration for a signal to travel from the Earth station to the satellite and back). Thus, if the satellite were placed in an orbit at three times the distance, it would take 800 km divided by 27,359 km/h, resulting in 29 hours, multiplied by 3, leading to 3 days 6 hours and 25 minutes.

Additional Information

  • Satellites can be categorized based on their orbital heights, but they can function at varying altitudes. Low Earth Orbit (LEO): 300 km - 1500 km above the Earth's surface.
  • Medium Earth Orbit (MEO): 1500 km - 36000 km.
  • Geosynchronous Orbit (GSO): approximately 36,000 km above the Earth's surface.
  • Geostationary Orbit (GEO): 35,790 km above the Earth's surface.
  • High Earth Orbit (HEO): above 36,000 km.

Additionally, there are specialized orbits used to classify satellites:

  • Molniya orbit, characterized by a 12-hour orbital period and an inclination of around 63°.
  • Sun-synchronous orbit, which has a specific angle and altitude that allows the satellite to cross the equator at the same local time.
  • Polar orbit, which takes the satellite over the poles

Learn More:

Satellite: ,

Details:

Class: Senior High

Subject: Physics

Keyword: Satellite

serg [3.5K]1 month ago
5 0
Kepler's third law, referred to as the law of harmonies, is used to calculate the orbital period and radius of a planet based on the dimensions and periods of another planet. This relationship is directly proportional to the square of the period and inversely proportional to the cube of the distance. Therefore, when the distance is tripled ((3D)^3), the period should increase to the square root of 27 times 5.20 times the initial period,
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A particular brand of gasoline has a density of 0.737 g/mLg/mL at 25 ∘C∘C. How many grams of this gasoline would fill a 14.9 gal
Yuliya22 [3333]

Answer:

The answer to your inquiry is Mass = 41230.7 g or 41.23 kg.

Explanation:

Data

Density = 0.737 g/ml

Mass = ?

Volume = 14.9 gal

1 gal = 3.78 l

Process

1.- Convert gallons to liters

1 gal ---------------- 3.78 l

14.9 gal ------------- x

x = 56.44 l

2.- Convert liters to milliliters

1 l ------------------- 1000 ml

56.44 l --------------- x

x = (56.44 x 1000) / 1

x = 56444 ml

3.- Calculate the mass

Formula

Density = \frac{mass}{volume}

Solving for mass

Mass = density x volume

Substituting values

Mass = 0.737 x 56444

Result

Mass = 41230.7 g or 41.23 kg.

3 0
3 months ago
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2
Ostrovityanka [3204]

The complete question is;

Block 1 sits on the floor with block 2 resting atop it. Block 3, which is stationary on a frictionless table, is attached to block 2 via a string that passes over a pulley depicted in the illustration below. Both the string and pulley have negligible mass.

Once block 1 is taken away without impacting block 2.

Derive an equation for the acceleration of block 3 considering arbitrary values for m3 and m2. Express your answer in terms of m3, m2, and relevant physical constants as needed.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Examining the attached illustration, by analyzing the free body diagram for block 3 and utilizing Newton's first law of motion, we reach the following formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is the mass of block 3

Simultaneously, doing the same for Block 2, the free body diagram yields the equation; (m2)g - T = (m2)a

Rearranging for T results in;

T = (m2)g - (m2)a - - - (eq 2)

where;

g represents acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is the mass of block 2

To deduce the acceleration, we will substitute (m3)a in place of T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0
1 month ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
Softa [3030]

Answer:

57.94°

Explanation:

We understand that the formula for flux is

\Phi =E\times S\times COS\Theta

where Ф represents flux

           E indicates electric field

           S denotes surface area

        θ signifies the angle between the electric field direction and the surface normal.

It is given that Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
2 months ago
Ability of the muscles to function effectively and efficiently without undue fatigue
Keith_Richards [3271]

Response:

Physical well-being

Clarification:

7 0
3 months ago
A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation
Ostrovityanka [3204]

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

6 0
2 months ago
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