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3 years ago

A satellite of mass M moves in a circular orbit of radius R at a constant speed v around the Earth which has mass ME.

Physics

1 answer:

Gwar [14]3 years ago

3 0

-- "Work" is the product of (force exerted on the object) x (distance the object moves in the direction of the force).

Since the orbit is a circle, the gravitational force toward the center is always perpendicular to the orbit. The object never moves in the direction of the force. If it did, it wouldn't be 'R' away from the center of the circle.

So the product of (gravitational force) x (distance in the direction of the force) is always zero.

-- Even if the orbit ISN't a circle . . . there are some parts of the orbit that aren't quite perpendicular to the gravitational force. If the satellite is traveling through one of those parts AND getting closer to the central body, then gravity is doing positive work on the satellite. If the satellite is traveling through one of those parts and getting FARTHER from the central body, then the satellite is the one doing positive work, and gravity is doing 'negative work'. The work done by gravity ... and the work done by the satellite ... is zero over a complete revolution, although not zero at every point.

This is exactly the definition of a "Conservative Force" ... a force that does zero work through one trip around any CLOSED path. Gravity is a conservative force, and so is the Electrostatic force.

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Why transparent object don't form a shadow

Pani-rosa [81]

Explanation:

Transparent objects do not form shadows. The light passes completely from the transparent objects thus these objects will not form shadow. ... In such objects, the light gets refracted thus, such objects forms shadow. The refraction is also the reason why we can see such objects.

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3 years ago

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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

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3 years ago

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string

For without breaking the string tension must be equal to the centripetal force

So $T=\frac{mv^2}{r}$

So $350=\frac{0.5\times v^2}{1}$

$v^2=700$

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0

3 years ago

A wire 1 mm in diameter is connected to one end of a wire of the same material 2 mm in diameter of twice the length. A voltage s

miskamm [114]

Answer:

T = 2 T₀

Explanation:

To answer this question let's write the expression for electrical conductivity

σ = n e2 τ / m*

The relationship with resistivity is

ρ = 1 /σ

Whereby the resistance

R = ρ L / A = 1 /σ L / A

We see that there is no explicit relationship between time and resistance, there is only a dependence on the life time (τ) that depends on the properties of the material, not on its diameter or length.

As also the average velocity or electron velocity of electrons is constant, the time to cross 2 mm in length is twice as long as the time to cross a mm in length

T = 2 T₀

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3 years ago

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sweet [91]

<h3>WATER</h3>

Explanation:

<h2>#CARRYINGTOLEARN:)</h2>

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3 years ago

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