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Comparative Planetology Determining Planet CharacteristicsPowerPoint Presentation

Comparative Planetology Determining Planet Characteristics

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The Apollo 8 spacecraft spent 20 hours in orbit around the moon. During this time, it completed 10 orbits. Assume the orbit was circular, with a radius of 1.86 x 106 m. The radius of the moon’s orbit around the earth is 3.8 x 108 m. The radius of the moon is 1.74 x 106 m.

The Apollo 8 spacecraft spent 20 hours in orbit around the moon. During this time, it completed 10 orbits. Assume the orbit was circular, with a radius of 1.86 x 106 m. The radius of the moon’s orbit around the earth is 3.8 x 108 m. The radius of the moon is 1.74 x 106 m.

Determining Planet Characteristics

Most of what is in this tutorial is a review of information contained in past tutorials

Please refer to the previous tutorials to reinforce the information contained in this tutorial

The Apollo 8 spacecraft spent 20 hours in orbit around the moon. During this time, it completed 10 orbits. Assume the orbit was circular, with a radius of 1.86 x 106 m. The radius of the moon’s orbit around the earth is 3.8 x 108 m. The radius of the moon is 1.74 x 106 m.

As viewed from earth, what is the angular diameter of the orbit of Apollo 8?

Radius of Moon’s Orbit

θ

Radius of Apollo 8 Orbit

The Apollo 8 spacecraft spent 20 hours in orbit around the moon. During this time, it completed 10 orbits. Assume the orbit was circular, with a radius of 1.86 x 106 m. The radius of the moon’s orbit around the earth is 3.8 x 108 m. The radius of the moon is 1.74 x 106 m.

As viewed from earth, what is the angular diameter of the orbit of Apollo 8?

Using a right triangle,

Tan θ = (Radius of Apollo 8 orbit) / (Radius of moon’s orbit)

= (1.86 x 106 m ) / ( 3.8 x 108 m ) = 0.489 x 10-2 = 0.00489

→ θ = 0.28o = 1009.5 arc sec.

The angular diameter of the orbit is, therefore,

θ = 0.56o = 2019 arc sec.

The Apollo 8 spacecraft spent 20 hours in orbit around the moon. During this time, it completed 10 orbits. Assume the orbit was circular, with a radius of 1.86 x 106 m. The radius of the moon’s orbit around the earth is 3.8 x 108 m. The radius of the moon is 1.74 x 106 m.

Based on the previous data, what is the mass of the moon?

Recall that

M = r v2 / G

Where v = 2πr / P is the speed of the Apollo 8 spacecraft, and M is the mass of the moon. See next slide for details.

The Apollo 8 spacecraft spent 20 hours in orbit around the moon. During this time, it completed 10 orbits. Assume the orbit was circular, with a radius of 1.86 x 106 m. The radius of the moon’s orbit around the earth is 3.8 x 108 m. The radius of the moon is 1.74 x 106 m. Based on the previous data, what is the mass of the moon?

Recall that

M = r v2 / G

From the Universal Law of Gravity and Newton’s Second Law,

Fon m due to M = G (M m) / (r2 ) = m a

Therefore, a = G M / r2

If the orbit is circular,

a = G M / r2 = v2 / r.

And therefore. M = r v2 / G, where M is the mass of the object that is the “provider” of the gravitational force on m (in this case, M is the mass of the moon.

m

r

M

Based on the previous data, what is the mass of the moon?

Recall that

M = r v2 / G

Where v = 2πr / P is the speed of the Apollo 8 spacecraft, and M is the mass of the moon.

The period of the orbit is 2 hours = 7200 seconds

So v = 2 (3.14) (1.86 x 106 m) / (7200 sec) = 1.6 x 103 m/sec

And therefore

MMoon = (1.86 x 106 m) (1.6 x 103 m/sec)2 / (6.67 x 10-11 N m2 / kg2 )

= 0.713 x 1023 kg = 7.13 x 1022 kg

Based on the previous data, what is the density of the moon?

DensityMoon = MMoon / VMoon

Where VMoon = Volume of the moon

= 4/3 π RMoon3

= 4/3 (3.14) (1.74 x 106 m)3

= 22 x 1018 m3 = 2.2 x 1019 m3

Therefore

DensityMoon = (7.13 x 1022 kg ) / (2.2 x 1019 m3 ) = 3.2 x 103 kg/m3

FYI – this is about 61% of the density of the earth.

The Apollo 8 spacecraft spent 20 hours in orbit around the moon. During this time, it completed 10 orbits. The radius of the moon’s orbit around the earth is 3.8 x 108 m. The radius of the moon is 1.74 x 106 m.

If the orbit of Apollo 8 was actually elliptical, with a semimajor axis of 1.86 x 106 m and an eccentricity of e =0.01, what is the closest the spacecraft would be to the surface of the moon, and what is the furthest it would be from the surface of the moon?

From our work on ellipses,

Distance of closest approach = a (1 – e) = 1.86 x 106 m (1 – 0.01) = 1.84 x 106 m

Farthest distance = a (1 + e) = 1.86 x 106 m (1 + 0.01) = 1.88 x 106 m

However, remember that these numbers are as measured from the CENTER of the moon, so subtracting the radius of the moon

Closest to the surface = 1.84 x 106 m – 1.74 x 106 m = 0.1 x 106 m = 1 x 105 m

Farthest from surface = 1.88 x 106 m – 1.74 x 106 m = 0.14 x 106 m = 1.4 x 105 m

Given that the actual mass of the moon is 7.35 x 10 moon. During this time, it completed 10 orbits. The radius of the moon’s orbit around the earth is 3.8 x 1022 kg (the previous rough calculation is pretty accurate), what is your weight on the moon? Assume your mass is 70 kg.

Your weight on the moon would be the same as the force of gravity on your body as produced by the moon. Recall that your weight on the moon would be given by

Won the Moon = m gMoon

Where gMoon is the acceleration due to gravity at the surface of the moon. Also recall that (using the Universal Law of Gravity),

gMoon = G MMoon / RMoon2

= (6.67 x 10-11 N m2 / kg2 ) (7.35 x 1022 kg) / (1.74 x 106 m)2

= 1.62 m/sec2

Therefore, on the moon, you would weigh

W = m gMoon = (70 kg) (1.62 m/sec2 ) = 113.3 N = 25 pounds.

By comparison, on the earth

W = m gEarth = (70 kg) (9.8 m/sec2 ) = 686 N = 156 pounds.

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