21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 13, Problem 10QP
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GMm
F,
r2
F(r = R)
The surface gravity g
of a body is. The acceleration due to gravity that an object
m
would feel on the surface of the body.
A. Show that the surface gravity of Earth is ge = 9.8 m/s².
B. Determine the surface gravity of the Sun.
C. Determine the surface gravity of the Sun when it becomes a red giant star, assuming
RG 1 AU. Use this answer to explain the significant mass loss rates observed in
these objects.
The first planets found in a pulsar system were found to have a mass of about
a.
1 Earth mass.
b.
3–4 Earth masses.
c.
20 Earth masses.
d.
100 Earth masses.
e.
zero, because planets have not been found around pulsars.
5. We observe an increase in brightness of a star that is 5.81×10¹⁹ m away. How long ago did the
actual increase in brightness take place, in years? Find the time that is required for light to reach
the Earth.
Chapter 13 Solutions
21st Century Astronomy
Ch. 13.1 - Prob. 13.1CYUCh. 13.2 - Prob. 13.2CYUCh. 13.3 - Prob. 13.3CYUCh. 13.4 - Prob. 13.4CYUCh. 13 - Prob. 1QPCh. 13 - Prob. 2QPCh. 13 - Prob. 3QPCh. 13 - Prob. 4QPCh. 13 - Prob. 5QPCh. 13 - Prob. 6QP
Ch. 13 - Prob. 7QPCh. 13 - Prob. 8QPCh. 13 - Prob. 9QPCh. 13 - Prob. 10QPCh. 13 - Prob. 11QPCh. 13 - Prob. 12QPCh. 13 - Prob. 13QPCh. 13 - Prob. 14QPCh. 13 - Prob. 15QPCh. 13 - Prob. 16QPCh. 13 - Prob. 17QPCh. 13 - Prob. 18QPCh. 13 - Prob. 19QPCh. 13 - Prob. 20QPCh. 13 - Prob. 21QPCh. 13 - Prob. 22QPCh. 13 - Prob. 23QPCh. 13 - Prob. 24QPCh. 13 - Prob. 25QPCh. 13 - Prob. 26QPCh. 13 - Prob. 27QPCh. 13 - Prob. 28QPCh. 13 - Prob. 29QPCh. 13 - Prob. 30QPCh. 13 - Prob. 31QPCh. 13 - Prob. 32QPCh. 13 - Prob. 33QPCh. 13 - Prob. 34QPCh. 13 - Prob. 35QPCh. 13 - Prob. 36QPCh. 13 - Prob. 37QPCh. 13 - Prob. 38QPCh. 13 - Prob. 39QPCh. 13 - Prob. 40QPCh. 13 - Prob. 41QPCh. 13 - Prob. 42QPCh. 13 - Prob. 43QPCh. 13 - Prob. 44QPCh. 13 - Prob. 45QP
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- I II E C F(O Compare D and E for brightness A. D>E B. E>D C. D=Earrow_forwardWhere does gold (the element) come from? A. It is produced during the supernova explosions of high-mass stars. B. It was produced during the Big Bang. C. It is produced by mass transfer in close binaries. D. It is produced during the late stages of fusion in low-mass stars.arrow_forward2. Kittel, Ch4-5. Surface temperature of the Earth. Calculate the temperature of the surface of the Earth on the assumption that as a black body in thermal equilibrium it reradiates as much thermal radiation as it receives from the Sun. Assume also that the surface of the Earth is at a constant temperature over the day-night cycle. Use T = 5800K; R₂ = 7×10¹⁰ cm; and the Earth-Sun distance of 1.5x10¹3 cm.arrow_forward
- 1- 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_forwardA helium flash a. is sudden and powerful. b. destroys the star. c. is a slow expulsion of helium gas. d. occurs in very low mass stars.arrow_forwardThe Schwarzschild radius of our sun is a. 3 km. b. 6 km. c. 9 km. d. 30 km.arrow_forward
- Because of the precession of the Earth’s axis, a. there are four seasons, spring, summer, fall, and winter. b. the Earth receives more solar radiation in the summer than in the winter. c. Polaris will not be the North Star in about 12,000 years. d. the lengths of the Earth’s days and nights vary throughout the year.arrow_forwardAn astronomical unit a.is the average distance from the Earth to the Sun. b.is about 150 million kilometres. c.is measured using radar signals bouncing off Venus. d.has units of distance. e.all of the mentioned choices. Galileo made many discoveries. Which one of these is not attributed to him? a.that the Moon has mountains, valleys and craters b.that the Sun has imperfections known as sunspots c.that four small points of light orbit the planet Jupiter d.the invention of the telescope e.that Venus shows a complete cycle of phases Which one of these statements describes why the planet Venus is much warmer than Earth? a.Venus has a denser and deeper atmosphere. b.It is closer to the Sun than Earth. c.It experienced a 'runaway greenhouse effect' long ago. d.It produces more carbon dioxide than is absorbed by the surface. e.all of the mentioned statementsarrow_forwardQUESTION 1 Estimate The Temperature For A Planet In Other Solar System (Questions 1-3) Let us assume scientists just discovered a planet orbiting a star in an extra-solar system. The star has a surface temperature Ts = 10000 Kelvins and a radius Sr = 1x109 meters. Scientists also measured the distance (D) between the star and the planet as D = 2 AU - 3.0x1011 meters. The solar power per unit area from the star's surface (Ps) can be calculated from the star's surface temperature Ts (10000 Kelvins) by the Stefen-Boltzman law Ps=0(Ts)4, where o is Stefen-Boltzman constant (5.67 x 10-8 Watt/meter2/Kelvin4). What is the solar power per unit area from the star's surface (Ps)? O Ps ~ 2.87 x 108 Watt/meter2 O Ps ~ 5.67 x 108 Watt/meter2 O O Ps ~ 2.87 x 10 Watt/meter2 Watt/meter² Ps ~ 5.67 x 10⁹ QUESTION 2 The solar power (Ps) decreases from the star's surface to the distance at the planet. Assuming the solar power per unit area at the distance of the planet as Pp, we have Pp=Ps(Sr/D)2, where…arrow_forward
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