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3 years ago

Explain how Galileo proved that his model was correct and that Ptolemy’s was not correct.

Physics

2 answers:

Molodets [167]3 years ago

6 0

Answer:

Written below.

Explanation:

Despite his many attempts, Galileo could not prove that the earth went around the sun. However, he was able to prove that the Ptolemaic model was incorrect, after he made telescopic observations of Venus. He discovered that Venus went through a full set of phases, like our moon. This could only happen if Venus went around the sun. In the Ptolemaic model, Venus does not go around the sun, and so would not go through all the phases. The diagram below shows how Venus would only go through crescent phases in the Ptolemaic model. (Recall that the phase of Venus will depend on the relative positions of the earth, Venus and the sun.)

In order to see a "full" Venus, the sun must be between the earth and Venus. A "gibbous" Venus is when the sun is sort of between the earth and Venus. When Venus is at its greatest elongation from the sun, then we would see it as a "half." When Venus was even closer to the earth, we would see it as a crescent. All these conditions happen only when Venus goes around the sun, as in the models by Copernicus and Brahe. In the Ptolemaic model, Venus is always between the earth and the sun, so would only show crescents. Even greatest elongation would not give a half Venus, because the triangle made between the sun, earth and Venus would not be a right triangle.

Galileo's studies of motion also led him to discover the basic idea of inertia. He finally figured out that if an object has a velocity, it will maintain that velocity without the need for any other forces. In fact, it takes a force in order to change a velocity. The reason why objects slow down unless they are pushed is that there is a frictional force that acts whenever two objects rub or slide across each other, and this frictional force is what causes things to slow down. As scientists finally began to accept Galileo's ideas of motion, then the big scientific reasons for rejecting a moving earth disappeared, leaving only the religious.

By this time Kepler had discovered that the planets actually orbitted the sun in an ellipse, but I am not sure why Galileo did not mention Kepler. It should also be noted that Brahe had proposed his own hybrid model of the solar system, with the sun orbiting the earth, and all the other planets orbiting the sun. Brahe's model did correctly explain the phases of Venus, and had the earth at rest in the center. For a while in the seventeenth century, the religious astronomers dumped Ptolemy, but clung to Brahe's as a possible model that kept the earth at rest. Science as a whole, however, had moved on and accepted the heliocentric model. Most scientists accepted the heliocentric theory not because there was proof that the earth moved, but because it was the "prettiest" model that explained the observations.

kkurt [141]3 years ago

5 0

Answer:

Despite his many attempts, Galileo could not prove that the earth went around the sun. However, he was able to prove that the Ptolemeic model was incorrect, after he made telescopic observations of Venus. He discovered that Venus went through a full set of phases, like our moon

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A projectile is launched at an angle of 30 and lands 20 s later at the same height as it was launched. (a) What is the initial s

Pavlova-9 [17]

Answer:

(a) 196 m/s

(b) 490 m

(c) 3394.82 m

(d) 2572.5 m

Explanation:

First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.

Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.

<u>Given:</u>

Time of flight = t = 20 s
Angle of the initial velocity of projectile with the horizontal = $\theta = 30^\circ$

<u>Assume:</u>

Initial velocity of the projectile = u
R = Range of the projectile during the time of flight
H = maximum height of the projectile
D = displacement of the projectile from the initial position at t = 15 s

Let us assume that the position from where the projectile was projected lies at origin.

Initial horizontal velocity of the projectile = $u\cos \theta$
Initial horizontal velocity of the projectile = $u\sin \theta$

Part (a):

During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.

$\therefore u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow u\sin \theta t=\dfrac{1}{2}(g)t^2\\\Rightarrow u=\dfrac{gt^2}{2\sin \theta t}\\\Rightarrow u=\dfrac{9.8\times 20^2}{2\sin 30^\circ \times 20}\\\Rightarrow u=196\ m/s$

Hence, the initial speed of the projectile is 196 m/s.

Part (b):

For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.

So, time taken to reach its maximum height will be equal to 10 s.

And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.

$\therefore H = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow H = 196\times \sin 30^\circ \times 10 + \dfrac{1}{2}\times(-9.8)\times 10^2\\ \Rightarrow H =490\ m$

Hence, the maximum altitude is 490 m.

Part (c):

Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.

So, the range is given by:

$R = u\cos \theta t\\\Rightarrow R = 196\times \cos 30^\circ \times 20\\\Rightarrow R =3394.82\ m$

Hence, the range of the projectile is 3394.82 m.

Part (d):

In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.

$\therefore h = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow h = 196\times \sin 30^\circ \times 15 + \dfrac{1}{2}\times(-9.8)\times 15^2\\ \Rightarrow h =367.5\ m\\r = u\cos \theta t\\\Rightarrow r = 196\times \cos 30^\circ \times 15\\\Rightarrow r =2546.11\ m\\\therefore D = \sqrt{r^2+h^2}\\\Rightarrow D = \sqrt{2546.11^2+367.5^2}\\\Rightarrow D =2572.5\ m$

Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.

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Answer:

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BUT since the question asked "would it pop out the other side?", I'm assuming it's talking about northern to southern hemisphere. so in that case it would pop out the other side since gravity makes things go downwards.

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