21ST CENTURY ASTRONOMY >CUSTOM<
6th Edition
ISBN: 9781324027836
Author: Kay
Publisher: NORTON
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Chapter 10, Problem 35QP
To determine
The distance from the sun where its brightness is same as the brightness of the full moon on Earth and the comparison of this distance with the semi-major axis Neptune’s orbit.
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Mercury's orbit ranges from 46 to 70 million km from the Sun, while Earth orbits at about 150 million km.
a. The Sun has a 30-arc-minute diameter viewed from Earth; what range of sizes does it have when viewed from Mercury?
When Mercury is 46 million km from the Sun, the Sun has a diameter of
When Mercury is 70 million km from the Sun, the Sun has a diameter of
arc-minutes.
arc-minutes.
b. At Mercury's orbital extremes, how many times stronger is the Sun's radiation on Mercury than on Earth?
At 46 million km, the Sun's radiation is
times stronger than on Earth.
At 70 million km, the Sun's radiation is
times stronger than on Earth.
EAn astronaut arrives on the planet Oceania and climbs to the top of a cliff overlooking the sea. The astronaut's eye is 100 m above
the sea level and he observes that the horizon in all directions appears to be at angle of 5 mrad below the local horizontal. What is
the radius of the planet Oceania at sea level? How far away is the horizon from the astronaut?
6000 km and 50 km
3600 km and 20 km
2000 km and 40 km
8000 km and 40 km
Why are all large celestial bodies (stars, planets, larger moons) very nearly spherical in shape?
a
because of the centrifugal force from the body's rotation
b
because of tidal forces
c
because gravity tries to pull every part of the celestial body to the center
d
because of the pressure from the heat in the body's core
Chapter 10 Solutions
21ST CENTURY ASTRONOMY >CUSTOM<
Ch. 10.1 - Prob. 10.1CYUCh. 10.2 - Prob. 10.2CYUCh. 10.3 - Prob. 10.3CYUCh. 10.4 - Prob. 10.4CYUCh. 10.5 - Prob. 10.5CYUCh. 10 - Prob. 1QPCh. 10 - Prob. 2QPCh. 10 - Prob. 3QPCh. 10 - Prob. 4QPCh. 10 - Prob. 5QP
Ch. 10 - Prob. 6QPCh. 10 - Prob. 7QPCh. 10 - Prob. 8QPCh. 10 - Prob. 9QPCh. 10 - Prob. 10QPCh. 10 - Prob. 11QPCh. 10 - Prob. 12QPCh. 10 - Prob. 13QPCh. 10 - Prob. 14QPCh. 10 - Prob. 15QPCh. 10 - Prob. 16QPCh. 10 - Prob. 17QPCh. 10 - Prob. 18QPCh. 10 - Prob. 19QPCh. 10 - Prob. 20QPCh. 10 - Prob. 21QPCh. 10 - Prob. 22QPCh. 10 - Prob. 23QPCh. 10 - Prob. 24QPCh. 10 - Prob. 25QPCh. 10 - Prob. 26QPCh. 10 - Prob. 27QPCh. 10 - Prob. 28QPCh. 10 - Prob. 29QPCh. 10 - Prob. 30QPCh. 10 - Prob. 31QPCh. 10 - Prob. 32QPCh. 10 - Prob. 33QPCh. 10 - Prob. 34QPCh. 10 - Prob. 35QPCh. 10 - Prob. 36QPCh. 10 - Prob. 37QPCh. 10 - Prob. 38QPCh. 10 - Prob. 39QPCh. 10 - Prob. 40QPCh. 10 - Prob. 41QPCh. 10 - Prob. 42QPCh. 10 - Prob. 43QPCh. 10 - Prob. 44QPCh. 10 - Prob. 45QP
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- 3.6 A spacecraft is approaching Venus with V = 10 km/s and b = 10,000 km. What will be the periapsis radius at Venus? Solution: 7266 km.arrow_forwardThe figure below shows the radial velocity of a star plotted as a function of time over the course of 20 days. Where is the planet in its orbit around the star when the star's radial velocity is 18 km/s? How do I determine this?arrow_forward1- MODIS is an Earth Observation sensor onboard TERRA spacecraft flying in a near-polar circular orbit with an orbital period of 98.8 minutes. The width of the swath imaged by MODIS is 2330 km. A- How many orbits does TERRA trace in one day? B- Assuming that the Earth rotates around its polar axis at a rate of 0.2618 rd/hr and that the equatorial radius is 6378 km, do two consecutive swaths of MODIS overlap at the equator? (hint: the length of an arc = angle in rd * radius) C- The radius of the latitude circle at 35 deg is 5224.5 km. Do two consecutive swaths of MODIS overlap at latitude 35 deg? 2- An aerostationary orbit for Mars is equivalent to a geostationary orbit for Earth. It is designed to enable a satellite in that orbit to image always the same surface of Mars. Calculate the altitude of an aerostationary orbit assuming that Mars is spherical, that its sidereal rotational period is 1.02595676 Earth days, its equatorial radius is 3389.50 km and its mass is…arrow_forward
- Light takes about 8.3 minutes to travel from the Sun to Earth. When it is closest, Neptune is 30.0 times farther from Earth than the Sun is. Make a prediction: Should light take more or less time to reach Earth from Neptune than from the Sun?arrow_forwardUse Kepler's 3rd Law and the small angle approximation. a) An object is located in the solar system at a distance from the Sun equal to 41 AU's . What is the objects orbital period? b) An object seen in a telescope has an angular diameter equivalent to 41 (in units of arc seconds). What is its linear diameter if the object is 250 million km from you? Draw a labeled diagram of this situation.arrow_forwardConsider the attached light curve for a transiting planet observed by the Kepler mission. If the host star is identical to the sun, what is the radius of this planet? Give your answer in terms of the radius of Jupiter. Brightness of Star Residual Flux 0.99 0.98 0.97 0.006 0.002 0.000 -8-881 -0.06 -0.04 -0.02 0.00 Time (days) → 0.02 0.04 0.06arrow_forward
- Earth is about 150 million kilometers from the Sun (1 Astronomical Unit, or AU), and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Using these two facts and the inverse square law for light, determine the apparent brightness that we would measure for the Sun if we were located at the following positions. a) At the orbit of Jupiter (780 million km from the Sun).arrow_forwardThe sun of galaxy X has a diameter of about 796,000 mi with a maximum distance from Planet X's surface of about 78,700,000 mi. Planet X's moon has a diameter of 2,731 mi. For a total solar eclipse to occur, the moon must pass between the sun and Planet X. The moon must also be close enough to Planet X for the moon's umbra (shadow) to reach the surface of Planet X. The maximum distance that the moon can be from Planet X and still have a total solar eclipse occur i (Round to the nearest thousand as needed )arrow_forwardRaising a number in scientific notation to a power is easy: (5 x 105)² = (5)² x (105)² = 5 x 5 x 105 x 105 = 25 x 10(5 × 2) = 25 x 1010 = 2.5 x 101¹1 Keeping this in mind, what is the volume of the sun in km³? The radius of the sun is about 7 x 105 km, and the volume of a sphere is 4/3 x Pi x R³. (Use 3.14 for Pi, and enter your answer with two decimal places). km³ 3 What is the average density of the Sun? Density = mass / volume. The mass of the sun is 2.0x10³0 kg. kg/km³arrow_forward
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