Question

The square of the orbital period of a planet is proportional to the cube of its distance from the Sun. This is expressed in the formula T^2=a^3, where T is time, measured in years, and a is distance, measured in astronomical units (1 astronomical unit is the mean distance of Earth from the Sun). Use this information to answers questions 7-9.

7) Express a as a function of T. Express T as a function of a.

8) Pluto's orbital period is approximately 247.9 times that of Earth's. Estimate Pluto's mean distance from the Sun.

9) Venus's mean distance from the Sun is approximately 72.3% that of Earth's. Estimate Venus's orbital period.

Answer #1

7). If T^{2}=a^{3} , then a = T^{2/3}
and T = a^{3/2}.

8).In case of Pluto, the orbital period say, T_{p} is
approximately 247.9 times that of Earth's, say T_{E}. Let
Pluto’s mean distance from the Sun be a_{p}. Then
T_{p}^{2} = a_{p}^{3} or, (247.9
T_{E})^{2} = a_{p}^{3} or,
61454.41(T_{E})^{2} = a_{p}^{3} or,
61454.41*1^{3} = a_{p}^{3} so that
a_{p} =(61454.41)^{1/3} = 39.46 astronomical units
( on rounding off to 2 decimal places).

9). Let Venus's orbital period be T_{v} and let its mean
distance from the Sun be a_{V} . Then
T_{V}^{2}=a_{v}^{3} =
(72.3/100)^{3}*1^{3} = 0.3777933067. Then
T_{v} =(0.3777933067)^{1/2} = 0.61. ( on rounding
off to 2 decimal places). Thus, Estimate Venus's orbital period is
0.61 times that of the Earth i.e. 224 days (approximately).

The planet Mercury has an orbital period of 87.97 days, and its
greatest distance (aphelion distance) from the Sun is 0.4667 AU
{where 1 AU " 1 Astronomical Unit = 1.496x1011 m}. The mass of the
planet Mercury is 3.302x1023 kg and the mass of the Sun is
1.988x1030 kg. (a) Calculate the eccentricity of planet Mercury’s
orbit around the Sun. ! (b) Calculate the highest speed of planet
Mercury as it orbits around the Sun?

Suppose we find a planet with an orbital period of 200 days
around a star with the same mass as the Sun, but only 75 percent as
luminous. What is the planet’s semi-major axis, in AU?
For the planet in the problem above, how many times more flux
does it receive, relative to Earth?
(If you find that the flux is less than that of Earth, express
your answer as a decimal. For example, if the planet receives half
as...

The planet Gliese 163 c, discovered in 2012, is a potentially
habitable world which is approxi- mately 7 times same mass as the
Earth, orbiting a small red dwarf star that is only 0.4 times mass
of the Sun. Completing its entire orbit in only a little over 25
days, Gliese 163 c orbits at a distance from its star at a speed of
1.78 times that of the speed of the Earth around the Sun.
Approximating that both the...

Kepler’s Laws of Planetary Motion:
1. A newly discovered exoplanetary system has a planet orbiting
its host star at 50 times the distance from the Earth to the sun.
If the orbital period of the planet is 40 Earth years, what is the
mass of the host star?
2. The mass of the Earth is approximately 6 × 1026
kg. If the moon takes 27 days to orbit the Earth, how far is the
moon from the Earth?
3. A...

Suppose you discover a planet around a Sun-like star. From
observations you find an orbital period of one Earth month. Find
the semi-major axis using one of Kepler’s laws.

Planet Yoyo is discovered to orbit around a star with 5 times
the mass of our Sun. It orbits at a distance of 7 AU in a circular
orbit. The orbital period is measured to be 8 years. Find:
1) The mass of the planet. Give the answer in both solar masses
and kilograms
2) The mass of the star in kilograms
3) The distance of the planets orbit in meters.
4) The force due to gravity between the planet...

Planet X rotates in the same manner as the earth, around an axis
through its north and south poles, and is perfectly spherical. An
astronaut who weighs 944.0 N on the earth weighs 918.0 N at the
north pole of Planet X and only 850.0 N at its equator. The
distance from the north pole to the equator is 1.882×104
km , measured along the surface of Planet X.
Part A) How long is the day on Planet X?
t...

Determine the emission temperature
of the planet Venus. You may assume
the following: the mean radius of
Venus’ orbit is 0.72 times that of
the Earth’s orbit; the solar flux So
decreases as the square of the distance
from the Sun and has a value of
1367Wm?2 at the mean Earth orbit;
Venus’ planetary albedo = 0.77.

Planet X rotates in the same manner as the earth, around an axis
through its north and south poles, and is perfectly spherical. An
astronaut who weighs 946.0 N on the earth weighs 920.0 N at the
north pole of Planet X and only 848.0 N at its equator. The
distance from the north pole to the equator is 1.889×104
km , measured along the surface of Planet X.
a) How long is the day on Planet X?
b) If...

The key observation of Venus was that it exhibited a ________
phase.
new
gibbous
quarter
crescent
A complete physical theory to explain the motion of the planets
was developed by ...
Kepler
Ptolemy
Newton
Galileo
What is the force of gravity (in Newtons) acting between the
Earth and a 5-kg bowling ball that is resting on the surface of the
Earth?
Virtually no one acknowledged Newton's work during his
lifetime.
True...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 10 minutes ago

asked 29 minutes ago

asked 33 minutes ago

asked 42 minutes ago

asked 42 minutes ago

asked 48 minutes ago

asked 48 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago