Find the orbital speed v of a satellite in a circular orbit of radius r around a planet of mass m.

Answers 3

Answer:v=\sqrt{\frac{Gm}{r}}

Explanation:

Given

Orbital speed is v

Mass of planet is m

Radius of circular orbit is r

suppose M is the mas of satellite then centripetal force on satellite is equal to the Gravitational Pull.

\frac{Mv^2}{r}=\frac{GMm}{r^2}

where G=gravitational constant

thus on solving we get

v=\sqrt{\frac{Gm}{r}}

The orbital speed v of a satellite in a circular orbit of radius r around a planet of mass m should be v = √Gm/r.

Calculation of the orbital speed:

Since

Orbital speed is v

Mass of planet is m

The radius of the circular orbit is r

Here we assume M is the mass of the satellite so the centripetal force on the satellite should be equivalent to the Gravitational Pull.

So,

Mv^2 /r = GMm/r^2

Here

G=gravitational constant

So, v = √Gm/r.

AI generated Answer

v = √ (GM/r) Where G is the universal gravitational constant, M is the mass of the planet, and r is the radius of the satellite's circular orbit. The orbital speed v of a satellite in a circular orbit of radius r around a planet of mass m can be calculated using the equation v = √ (GM/r). The equation uses the universal gravitational constant G, the mass of the planet M, and the radius of the satellite's circular orbit, r, to calculate the orbital speed of the satellite, v.

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What is the difference? StartFraction x Over x squared + 3 x + 2 EndFraction minus StartFraction 1 Over (x + 2) (x + 1) EndFraction options A)StartFraction x minus 1 Over 6 x + 4 EndFraction B)StartFraction negative 1 Over 4 x + 2 EndFraction C)StartFraction 1 Over x + 2 EndFraction D)StartFraction x minus 1 Over x squared + 3 x + 2 EndFraction

Answer:

D

Step-by-step explanation:

x/(x²+3x+2) - 1/[(x+2)(x+1)]

x² + 3x + 2 = x² + 2x + x + 2

= x(x + 2) + (x + 2)

= (x + 2)(x + 1)

= x/[(x+2)(x+1)] - 1/[(x+2)(x+1)]

= (x-1)/[(x+2)(x+1)]

= (x-1)/(x²+3x+2)

The difference of rational polynomial function is \frac{x-1}{x^{2} +3x+2}

Given expression is,

          \frac{x}{x^{2} +3x+2}-\frac{1}{(x+2)(x+1)}

(x+2)(x+1)=x^{2} +2x+x+2=x^{2} +3x+2

Given expression can be written as,

       \frac{x}{x^{2} +3x+2}-\frac{1}{x^{2} +3x+2} =\frac{x-1}{x^{2} +3x+2}

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diameter of the trachea is 18mm, the air flow velocity out of the trachea is 20cm/s and the density of air is 1.225kg/m3. Also assume that lung volume is decreasing at a rate of 100mL/s.

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

Answer:

Time rate = 0.010026 kg/m³.s

Explanation:

We are given the following;

Total lung capacity; V = 6000mL = 6 x 10^(-3) m³

diameter of the trachea is 18mm = 0.018m

Air density; ρ = 1.225kg/m³

Velocity; v = 20cm/s = 0.20m/s

From the question, lung volume is decreasing at a rate of 100mL/s.

Thus, dv /dt = -100mL = -0.0001 m³/s

Area of trachea; A = πD²/4

A = (π x 0.018²)/4 = 2.5447 x 10^(-4) m²

Now, let's set up the equation.

-ρAv = (d/dt)(ρV) = V(dρ/dt) + ρ(dv/dt)

Thus,

Plugging in the relevant values to get;

-[1.225 x 2.5447 x 10^(-4) x 0.20] = 6 x 10^(-3)(dρ/dt) + (1.225 x -0.0001)

So,

-0.62345 x 10^(-4) = 6x10^(-3)(dρ/dt) - (1.225 x 10^(-4))

6x10^(-3)(dρ/dt) = (1.225 x 10^(-4)) - 0.62345 x 10^(-4)

6x10^(-3)(dρ/dt) = 0.60155 x 10^(-4)

(dρ/dt) = [0.60155 x 10^(-4)] /(6x10^(-3)) = 0.10026 x 10^(-1) = 0.010026 kg/m³.s

ρ

PART B: Which of the following quotes best supports the answer to Part A? A “Everyone, like myself, was probably on the verge of fetching a rope, or asking where we could find a ladder, but then we looked around at each other and it was decided.” ( Paragraph 2) B “I remember that we were still full of games and laughter when we called down to him. He heard us shouting while we were playing” ( Paragraph 3) C “At first afraid to disobey the voice from the man in the well, we turned around and actually began to walk toward the nearest house, which was Arthur’s. But along the way we slowed down” ( Paragraph 5) D “As we were rising quietly and looking at the dark shadows of the trees we had to move through to reach our homes, he said, ‘Why didn’t you tell anyone?’ He coughed. ‘Didn’t you want to tell anyone?’” ( Paragraph 56)

Answer: question to part a + story and I’ll help.

Explanation:

"My furniture, part of which i made myself... consisted of a bed, a table, a desk, three chairs, and looking-glass three inches in diameter." What is the meaning of "looking glass" in this sentence from the first paragraph? A) a mirror B) a chandelier C) a window pane D) a camera lens
A mirror which you look into
A mirror because your “looking at it”

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