How to Draw a Circle in Ps7

Trouble #1: Bianca is a moon of Uranus. Bianca has a circular orbit whose radius is 5.92 x 10^four km, and its orbital period is 0.435 days.

a) Calculate Bianca's speed in its orbit effectually Uranus.

b) From this information, calculate the mass of Uranus.

Solution: Speed is defined as distance per fourth dimension, e.thousand., "miles per hour," "meters per second," etc. In this trouble (every bit in the Exam!) we have both a distance and a fourth dimension, so computing the speed does non require some fancy formula.

Allow's start with a uncomplicated diagram of the situation:

Here, the moon Bianca orbits in a circumvolve around Uranus. How far does it travel to complete one lap around Uranus? Not the radius of its circular path, but the circumference, which is

• circumference = ii pi R
• = 2 ten 3.14 10 (5.92 x 10^four km)
• = 3.72 x 10⁵ km

Now this is a perfectly expert distance in perfectly good units, just since I know what'due south coming later, I'k going to catechumen information technology now into the more standard unit of meters:

• 3.72 ten 10^v km ten (1000 m/1 km)
• = 3.72 ten 10^viii thousand

At present, how long does it take for Bianca to complete one lap around Uranus? That'southward the menstruation, which is:

• period = 0.435 days
• = 0.435 days ten (24 hours/ 1 twenty-four hour period) x (60 minutes/ 1 hour) x (lx seconds/ 1 infinitesimal)
• = iii.76 x 10⁴ seconds

OK, and so at present we have the distance traveled in one lap and the fourth dimension information technology takes to travel this distance. And then,

• speed = distance/time
• = 3.72 ten 10⁸ g / iii.76 x 10⁴ seconds
• = 9.ix x 10³ one thousand/s, or 9900 m/s

At present for part b), where nosotros need to notice the mass of Uranus. We have the speed from role a) above, and we know that Bianca travels in a circular orbit. From these two pieces of information, we can summate the dispatch Bianca must feel, since, for any object traveling in a circle,

• dispatch = speed^two/radius
• = (9900 k/south)²/(5.92 x 10⁷ m)
• = 1.65 g/s^two

(Notation that I converted the radius to meters by 5.92 x 10⁴ km ten (1000 m/ 1 km) = 5.92 ten 10^vii m.)

This is the dispatch that Bianca must experience in club to travel in a circumvolve, and that dispatch must be acquired past the gravity of Uranus (there actually isn't any other choice). Newton told usa that the acceleration due to gravity is:

• acceleration = Thou x Mass (of planet)/ distance²

Here, discover that the mass involved is not the mass of Bianca, only the mass of the object responsible for the gravity (i.e., Uranus). We can rearrange this expression every bit follows and insert the dispatch we calculated for Bianca in its orbit, :

• Mass = acceleration x distance²/ G
• = i.65 one thousand/s² 10 (five.92 x x⁷ g)² / six.67 10 10^-11m³/kg s²
• = 8.67 x 10²⁵ kg

Now the units exercise in fact work out, but only if you used appropriate units for the acceleration and distance so that they cancel properly with the units for M. That's why it's of import in general to summate parameters in "standard" units of meters, kilograms, and seconds.

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Problem #ii: Mercury'due south atomic number 26 cadre occupies 40% of the volume of the planet, merely accounts for 60% of its total mass. Using the mass and radius of Mercury in the back of your textbook, summate the boilerplate density of the other part of Mercury, that is, the rocky outer layers.

Solution: Density is mass per volume, and so here we just need to figure out the mass and book of the outer part of Mercury. Since we're given the mass and volume of Mercury'southward inner iron cadre, we really simply need to subtract these values from the total mass and book of Mercury to get the value nosotros demand.

Looking in the back of our textbook, I observe that radius of Mercury is 2439 km, or,

• 2439 km x ( yard thou/1 km) = 2.439 x 10⁶ m

From this value, I can calculate Mercury's total volume (bold it'due south a sphere),

• volume = 4/three 10 pi x radius³
• = 1.33 10 3.14 ten (2.439 ten 10⁶ one thousand)ⁱⁱⁱ
• = 6.07 x x¹⁹ mⁱⁱⁱ

Now the book of Mercury's iron core is xl% of its total book, so the volume of what'south left must be 60% of Mercury's total book, so

• volume of outer part = 0.6 x 6.07 x x^xix one thousand³
• = iii.64 x x^nineteen yard³

Now, the mass of Mercury is 3.3 x 10²³ kg (from the back of our text), and the mass of its atomic number 26 core is sixty% of the total mass, so the mass of what's left must be 40% of Mercury's total mass,

• mass of outer part = 0.4 10 3.iii x x²³ kg
• = 1.32 10 10²³ kg

Now, density is just mass per volume, or

• density = mass/volume
• = ane.32 ten x²³ kg /iii.64 x ten^nineteen kⁱⁱⁱ
• = 3.63 ten 10³ kg/km³, or 3630 kg/km³

This is approximately the density of rocks in the Earth's chaff, which makes sense since we recall the outer office of Mercury has a similar rocky chaff.

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Problem #iii: The maximum elongation of Venus (that'due south the largest angular separation betwixt Venus and the Lord's day, viewed from Earth) is 46 degrees.

a) Describe a diagram indicating the position of the Sun and Earth, the orbit of Venus, and its location when it appears at its maximum elongation.

b) From this diagram (and the noesis that the Earth-Lord's day distance is one A.U.), calculate the radius of Venus' circular orbit.

Solution: The largest angle betwixt the direction to the Sun and the direction to Venus will occur for the post-obit geometry,

where the line from the World to Venus just touches the circle that defines Venus's orbit (mathematicians would say that this line is "tangent" to the circle). Nosotros can then make a triangle with vertices at the Earth, Venus, and the Dominicus,

where the angle at Venus is a right bending (i.e., 90 degrees). Why this one and non the one at the Sun? Look at the flick. If the bending at the Sun were a right bending, Venus would be at the "superlative" of its orbit in the flick. Only conspicuously, the elongation is larger when Venus'south just a little bit closer to the Earth than when information technology'due south at the top. And so the angle at the Dominicus must exist slightly smaller than 90 degrees.

The angle at Venus is 90 degrees because the tangent line is e'er perpendicular to the radius, as drawn at left beneath:

If the line weren't perpendicular to the radius line, it would "dig in" to the circumvolve and wouldn't be tangent (see other two figures in a higher place).

So, from the geometry of the Sun-Venus-Earth triangle, and knowing SOHCATOA, we get that

Sin(elongation) = Sun-Venus distance/Sun-Earth distance = Sun-Venus distance/ ane A.U.

and then,

Sun-Merc altitude = 1 A.U. x Sin(46 degrees) = 0.72 A.U.

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Source: https://www.eg.bucknell.edu/physics/astronomy/astr101/prob_sets/ps7_soln.html