A multiple-choice quiz
by jonnowales.
Estimated time: 6 mins.

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Quiz Answer Key and Fun Facts

Answer:
**Parallax**

There is a very nice algebraic expression that relates distance from the Earth to a particular star, d, and the angle, p, between the line from the Sun to the star and the line from the Earth to the star. This is essentially saying that a star is seen in one place on the celestial map when viewed from the sun but upon viewing that same star from Earth the star appears to have moved. This "apparent" movement of the star relative to other stars is called stellar parallax and the equation that defines it is:

p = 1 / d

The parallax angle, p, is measured in a unit called arcseconds and the distance from the observer to the star is measured in parsecs. This equation also shows us that the stars that appear to move the most, id est those with the greatest parallax angle, are the stars closest to the observer.

There is a very nice algebraic expression that relates distance from the Earth to a particular star, d, and the angle, p, between the line from the Sun to the star and the line from the Earth to the star. This is essentially saying that a star is seen in one place on the celestial map when viewed from the sun but upon viewing that same star from Earth the star appears to have moved. This "apparent" movement of the star relative to other stars is called stellar parallax and the equation that defines it is:

p = 1 / d

The parallax angle, p, is measured in a unit called arcseconds and the distance from the observer to the star is measured in parsecs. This equation also shows us that the stars that appear to move the most, id est those with the greatest parallax angle, are the stars closest to the observer.

Answer:
**Bolometer**

The luminosity of the sun, as detected by a bolometer, can be calculated by what is an ostensibly complicated equation but when one digs to find the physics behind the maths, it isn't so bad. The luminosity of the sun is said to be the product of the flux of an object's surface (the amount of energy emitted from a certain area per second) and the area of that surface. This is more easily seen in equation form:

Luminosity = Flux x Area

The flux is dependent upon temperature such that:

Flux = constant (sigma) x temperature to the fourth power.

This shows that if the temperature of the sun in a certain area doubles, then the flux increases by 2^4 = 2 x 2 x 2 x 2 = 16 times! The one assumption (apart from the issue of blackbody radiation) made when making a rough calculation of the luminosity of the sun is that we take the sun as a perfect sphere. The area of a perfect sphere is (4)*(pi)*(R^2), where 'R' is the radius of the sun.

After all this we arrive at the equation for luminosity, phew!

Luminosity = (4)*(pi)*(R^2)*(sigma)*(T^4)

The units of luminosity are (J/s) which means that luminosity is the amount of energy emitted from an object per unit time.

The luminosity of the sun, as detected by a bolometer, can be calculated by what is an ostensibly complicated equation but when one digs to find the physics behind the maths, it isn't so bad. The luminosity of the sun is said to be the product of the flux of an object's surface (the amount of energy emitted from a certain area per second) and the area of that surface. This is more easily seen in equation form:

Luminosity = Flux x Area

The flux is dependent upon temperature such that:

Flux = constant (sigma) x temperature to the fourth power.

This shows that if the temperature of the sun in a certain area doubles, then the flux increases by 2^4 = 2 x 2 x 2 x 2 = 16 times! The one assumption (apart from the issue of blackbody radiation) made when making a rough calculation of the luminosity of the sun is that we take the sun as a perfect sphere. The area of a perfect sphere is (4)*(pi)*(R^2), where 'R' is the radius of the sun.

After all this we arrive at the equation for luminosity, phew!

Luminosity = (4)*(pi)*(R^2)*(sigma)*(T^4)

The units of luminosity are (J/s) which means that luminosity is the amount of energy emitted from an object per unit time.

Answer:
**Radius & Temperature**

The equation for the luminosity of any spherical object (or assumed spherical object) is 'L = (4)*(pi)*(R^2)*(sigma)*(T^4)'; where 'R' is the radius of the spherical object, 'T' is the object's temperature and 'sigma' is the Stefan-Boltzmann constant.

When taking a ratio of the luminosities of two stars the numbers and constants will cancel each other out which will leave just 'L = (R^2)*(T^4)'. Thus the luminosity can be seen to be dependent on both the radius of the object and that object's temperature! Also to be noted is that a change in temperature will have a greater effect on luminosity than a change in radius of a similar magnitude.

The equation for the luminosity of any spherical object (or assumed spherical object) is 'L = (4)*(pi)*(R^2)*(sigma)*(T^4)'; where 'R' is the radius of the spherical object, 'T' is the object's temperature and 'sigma' is the Stefan-Boltzmann constant.

When taking a ratio of the luminosities of two stars the numbers and constants will cancel each other out which will leave just 'L = (R^2)*(T^4)'. Thus the luminosity can be seen to be dependent on both the radius of the object and that object's temperature! Also to be noted is that a change in temperature will have a greater effect on luminosity than a change in radius of a similar magnitude.

Answer:
**Main Sequence**

The main sequence is the part of stellar evolution where the star is most stable. The stability of the star is defined in terms of hydrostatic equilibrium, where the outward forces match the inward forces acting on the star. The inward forces are dominated by gravity whereby the force is trying to collapse the star as everything is attracted to the centre of the body. This is matched by the pressure exerted outwards from the core due to nuclear fusion processes. When this equilibrium is lost (due to all materials that are able to undergo fusion being used up) then the star will leave the stable main sequence and begin the process of stellar death. The larger the mass of the star, the shorter the time spent on the main sequence.

When low mass stars like our sun leave the main sequence they will usually end up as a rather lifeless, dull white dwarf. However, high mass stars have a far more interesting end and they literally go out with a bang! Sometimes the end of a high mass star's life will result in a supernova (a huge explosion) whilst others will just collapse into a black hole!

The main sequence is the part of stellar evolution where the star is most stable. The stability of the star is defined in terms of hydrostatic equilibrium, where the outward forces match the inward forces acting on the star. The inward forces are dominated by gravity whereby the force is trying to collapse the star as everything is attracted to the centre of the body. This is matched by the pressure exerted outwards from the core due to nuclear fusion processes. When this equilibrium is lost (due to all materials that are able to undergo fusion being used up) then the star will leave the stable main sequence and begin the process of stellar death. The larger the mass of the star, the shorter the time spent on the main sequence.

When low mass stars like our sun leave the main sequence they will usually end up as a rather lifeless, dull white dwarf. However, high mass stars have a far more interesting end and they literally go out with a bang! Sometimes the end of a high mass star's life will result in a supernova (a huge explosion) whilst others will just collapse into a black hole!

Answer:
**Apparent **

Hopefully some linguistic knowledge helped you out there as what you are seeing in the night sky is how the star "appears" to you. The star you are seeing from your point of observation on planet Earth however doesn't give us the whole story.

Due to the massive distances involved in astrophysics and astronomy, stars will not appear to us the way they would to a planet that was say two or three astronomical units away. This is taken into account with regards magnitude (brightness) because the magnitude of the star we see is affected by the distance the star is from us. So to determine the absolute magnitude of the star we need to play around with some equations and astrophysicists came up with this:

m - M = (5*log(d)) - 5

Where 'm' is the apparent magnitude, 'M' is the absolute magnitude and 'd' is the distance from the star to the observer in parsecs. Essentially the absolute magnitude is a standardisation of the magnitudes of stars by evaluating their intrinsic brightness at a distance of ten parsecs from the observer. By eliminating the affect of distance a fairer comparison between the magnitudes of stars can be made.

The part of the above equation 'm - M' is known as the distance modulus which is the measure of the difference between absolute and apparent magnitudes.

Hopefully some linguistic knowledge helped you out there as what you are seeing in the night sky is how the star "appears" to you. The star you are seeing from your point of observation on planet Earth however doesn't give us the whole story.

Due to the massive distances involved in astrophysics and astronomy, stars will not appear to us the way they would to a planet that was say two or three astronomical units away. This is taken into account with regards magnitude (brightness) because the magnitude of the star we see is affected by the distance the star is from us. So to determine the absolute magnitude of the star we need to play around with some equations and astrophysicists came up with this:

m - M = (5*log(d)) - 5

Where 'm' is the apparent magnitude, 'M' is the absolute magnitude and 'd' is the distance from the star to the observer in parsecs. Essentially the absolute magnitude is a standardisation of the magnitudes of stars by evaluating their intrinsic brightness at a distance of ten parsecs from the observer. By eliminating the affect of distance a fairer comparison between the magnitudes of stars can be made.

The part of the above equation 'm - M' is known as the distance modulus which is the measure of the difference between absolute and apparent magnitudes.

Answer:
**Distance**

A perfect example of why astrophysicists and astronomers are rather reluctant to give up their unique set of units is the concept of distance in the cosmos. Is a metre really an appropriate unit of distance when analysing outer space? Considering that the Earth is 93 million miles from the sun and that that distance (the astronomical unit (AU)) is not really that far in terms of the cosmos, a metre tends to insignificance.

A light year is a measure of distance, not time as is commonly thought. In fact it is the distance travelled by light in one year. Considering that light travels in a vacuum at approximately 300,000,000 metres per second and that there are 31,556,926 seconds in a year, then light travels 9,467,077,800,000,000 metres in one year. That number is ridiculous and would be a nightmare to manage in calculations involving multiples of the light year. If you are an astronomer and you like that number, then maybe the metre is the unit for you! If you are anything like me then you will probably work with the light year instead. My calculator certainly didn't appreciate sending me that hideous number!

A perfect example of why astrophysicists and astronomers are rather reluctant to give up their unique set of units is the concept of distance in the cosmos. Is a metre really an appropriate unit of distance when analysing outer space? Considering that the Earth is 93 million miles from the sun and that that distance (the astronomical unit (AU)) is not really that far in terms of the cosmos, a metre tends to insignificance.

A light year is a measure of distance, not time as is commonly thought. In fact it is the distance travelled by light in one year. Considering that light travels in a vacuum at approximately 300,000,000 metres per second and that there are 31,556,926 seconds in a year, then light travels 9,467,077,800,000,000 metres in one year. That number is ridiculous and would be a nightmare to manage in calculations involving multiples of the light year. If you are an astronomer and you like that number, then maybe the metre is the unit for you! If you are anything like me then you will probably work with the light year instead. My calculator certainly didn't appreciate sending me that hideous number!

Answer:
**H-R Diagram**

The H-R diagram is so named after the two scientists who came up with it - Dane, Ejnar Hertzsprung (H) and American, Henry Norris Russell (R). The Hertzsprung-Russell diagram is a plot of the logarithm of luminosity against the logarithm of temperature. By noting that the luminosity of a star is 'L = (4)*(pi)*(R^2)*(sigma)*(T^4)', (R = radius, T = temperature, sigma = Stefan-Boltzmann constant), it is possible to determine the slope upon which stars of the same radius will lie by using the rules of logarithms:

log(L) = 4*log(T) + 2*log(R) + log(4*pi*sigma)

The important part of the above equation is 'log(L) = 4*log(T) + 2*log(R)' because this is in the form 'y = mx + c' which is the equation for a straight line. The slope here is four which means that for stars that are perfect radiators (blackbodies) and of the same radius, the stars will lie on a slope of 4 on the Hertzsprung-Russell diagram. An amendment must be made to take into account the fact that the log of temperature decreases as one goes further along the positive x-axis, so the slope is -4. This is just one example of the use of the H-R diagram in determining where stars are in the stellar evolutionary cycles in terms of their physical characteristics.

The H-R diagram is so named after the two scientists who came up with it - Dane, Ejnar Hertzsprung (H) and American, Henry Norris Russell (R). The Hertzsprung-Russell diagram is a plot of the logarithm of luminosity against the logarithm of temperature. By noting that the luminosity of a star is 'L = (4)*(pi)*(R^2)*(sigma)*(T^4)', (R = radius, T = temperature, sigma = Stefan-Boltzmann constant), it is possible to determine the slope upon which stars of the same radius will lie by using the rules of logarithms:

log(L) = 4*log(T) + 2*log(R) + log(4*pi*sigma)

The important part of the above equation is 'log(L) = 4*log(T) + 2*log(R)' because this is in the form 'y = mx + c' which is the equation for a straight line. The slope here is four which means that for stars that are perfect radiators (blackbodies) and of the same radius, the stars will lie on a slope of 4 on the Hertzsprung-Russell diagram. An amendment must be made to take into account the fact that the log of temperature decreases as one goes further along the positive x-axis, so the slope is -4. This is just one example of the use of the H-R diagram in determining where stars are in the stellar evolutionary cycles in terms of their physical characteristics.

Answer:
**Hubble**

What I love about algebra is that it is so much easier to write down relationships in physics in terms of a few letters than it is to describe them in word form. Hubble's law is shown algebraically below:

v = H0 * d

Where 'v' is the recessional velocity, 'H0' is the Hubble constant and 'd' is the distance of the galaxy from Earth (or whatever the point of observation is).

The recessional velocity mentioned above is the velocity with which a galaxy moves away from the Earth. In the lead up to discovering that galaxies are moving away from us and that the Universe is expanding, it was noticed that there was a phenomenon known as redshift. A redshift is seen when light is travelling away from the observer. Essentially this just means that the light appears to shift towards the red end of the visible spectrum. If the light was moving towards the observer, the light would shift to the blue end of the visible spectrum.

For me, one of the incredible things about physics is that a simple equation such as 'v = H0 * d' can accurately (the whole issue of uncertainties aside) explain such fundamental events from the interactions in the atom to the recessional velocities of galaxies huge distances away. Surely there is no reason why physics can't be considered cool! I think it is cool but I am a bit of a geek!

What I love about algebra is that it is so much easier to write down relationships in physics in terms of a few letters than it is to describe them in word form. Hubble's law is shown algebraically below:

v = H0 * d

Where 'v' is the recessional velocity, 'H0' is the Hubble constant and 'd' is the distance of the galaxy from Earth (or whatever the point of observation is).

The recessional velocity mentioned above is the velocity with which a galaxy moves away from the Earth. In the lead up to discovering that galaxies are moving away from us and that the Universe is expanding, it was noticed that there was a phenomenon known as redshift. A redshift is seen when light is travelling away from the observer. Essentially this just means that the light appears to shift towards the red end of the visible spectrum. If the light was moving towards the observer, the light would shift to the blue end of the visible spectrum.

For me, one of the incredible things about physics is that a simple equation such as 'v = H0 * d' can accurately (the whole issue of uncertainties aside) explain such fundamental events from the interactions in the atom to the recessional velocities of galaxies huge distances away. Surely there is no reason why physics can't be considered cool! I think it is cool but I am a bit of a geek!

Answer:
**The rotational velocity should decrease**

Kepler's Third Law, 'p^2 = a^3'*, states that "the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." (en.wikipedia.org)

Isaac Newton modified Kepler's law by working out what the constant of proportionality in the equation was. The constant would be a bit messy to write on FunTrivia so suffice to say it involves pi, the universal gravitation constant, G, and masses. By analysing this new equation it can be simplified to the point that a relationship between rotational velocity, v, and the distance of a star from the galactic centre, r, becomes apparent. The relationship is that the rotational velocity of stars should *decrease* in an inversely proportionate manner with the square root of the radius: 'v is proportional to r^-(1/2)'.

However, the observations made by astronomers such as Vera Rubin showed that this just does not happen, the velocities actually stay approximately the same. This was astonishing as either there is something wrong with Newton's laws of gravitation (surely unthinkable!) or there is a heck of a lot of mass that we are not aware of - "dark matter"!

*Please note that I have said 'p^2 = a^3' in place of p^2 is directly proportional to a^3 for the purpose of brevity.

Kepler's Third Law, 'p^2 = a^3'*, states that "the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." (en.wikipedia.org)

Isaac Newton modified Kepler's law by working out what the constant of proportionality in the equation was. The constant would be a bit messy to write on FunTrivia so suffice to say it involves pi, the universal gravitation constant, G, and masses. By analysing this new equation it can be simplified to the point that a relationship between rotational velocity, v, and the distance of a star from the galactic centre, r, becomes apparent. The relationship is that the rotational velocity of stars should *decrease* in an inversely proportionate manner with the square root of the radius: 'v is proportional to r^-(1/2)'.

However, the observations made by astronomers such as Vera Rubin showed that this just does not happen, the velocities actually stay approximately the same. This was astonishing as either there is something wrong with Newton's laws of gravitation (surely unthinkable!) or there is a heck of a lot of mass that we are not aware of - "dark matter"!

*Please note that I have said 'p^2 = a^3' in place of p^2 is directly proportional to a^3 for the purpose of brevity.

Answer:
**Black Hole**

The Schwarzschild radius (Rs) is quantified in terms of the universal gravitation constant, G, the mass of the orbiting object, m, and the speed of light in a vacuum, c. The equation is:

Rs = (2*G*m)/(c^2)

The value of the Schwarzschild radius for a black hole is equivalent to the event horizon, the point of no return! Once an object, even one with a velocity as great as light, passes the event horizon there is no coming back and the object, well...we don't really know what happens. It is likely that it involves the object being ripped apart into its constituent atoms. How pleasant!

The Schwarzschild radius (Rs) is quantified in terms of the universal gravitation constant, G, the mass of the orbiting object, m, and the speed of light in a vacuum, c. The equation is:

Rs = (2*G*m)/(c^2)

The value of the Schwarzschild radius for a black hole is equivalent to the event horizon, the point of no return! Once an object, even one with a velocity as great as light, passes the event horizon there is no coming back and the object, well...we don't really know what happens. It is likely that it involves the object being ripped apart into its constituent atoms. How pleasant!

This quiz was reviewed by FunTrivia editor crisw before going online.

Any errors found in FunTrivia content are routinely corrected through our feedback system.

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