Astronomy Assignment

Name: _______________________ DATE: ____________________

Lab 8 - PROPERTIES OF PLANETARY NEBULAE

GOALS: Estimate the size and age of a planetary nebula

Quantify the mass of planetary nebulae returned to ISM

The NASA Hubble Space Telescope has captured the sharpest view yet

of the most famous of all planetary nebulae: the Ring Nebula (M57). The colors

are approximately true colors. The color image was assembled from three black-

and-white photos taken through different color filters with the Hubble

telescope's Wide Field Planetary Camera 2. Blue isolates emission from very hot

helium, which is located primarily close to the hot central star. Green represents

ionized oxygen, which is located farther from the star. Red shows ionized

nitrogen, which is radiated from the coolest gas, located farthest from the star.

The gradations of color illustrate how the gas glows because it is bathed in

ultraviolet radiation from the remnant central star, whose surface temperature

is a white-hot 216,000 degrees Fahrenheit (120,000 degrees Celsius). Most Sun-

like stars become planetary nebulae at the end of their lives. Once a star

consumes all of its hydrogen, the nuclear fuel that makes it shine, it expands to

a red giant. The bloated star then expels its outer layers, exposing its hot core.

Ultraviolet radiation from the core illuminates the discarded material, making it

glow. The smoldering core, called a white dwarf, is the tiny white dot in the

center of the Ring Nebula.

In this experiment, you will start by examining M57 to learn about

planetary nebulae and the role they play in enriching the interstellar medium.

Then, you will repeat same steps another no less famous planetary nebula

NGC7293 – known as the Helix Nebula.

To begin, view and take notes for the Hubble images of M57 and NGC7293

found at:

M57 https://hubblesite.org/contents/media/images/1999/01/748-Image.html

NGC7293 https://hubblesite.org/contents/media/images/2004/32/3911-Image.html

PART 1: THE SIZE AND NATURE OF M57, THE RING NEBULA

In the Hubble image of M57, the brighter regions are places where the density of

the nebula is high because there is more material present to create emission.

The negative image of M57 is provided at the end of this document, to be used

for this exercise

1. Does the nebula look like it is evenly filled, or does it look like a bubble? How

can you tell?

2. Use a ruler to measure the size of the nebula in centimeters. Measure the size

of the nebula along its longest axis. Then, divide this number by two to get the

radius of the nebula. Record the length of the longest axis and radius (in

centimeters to nearest millimeter)

Length (cm) = ___________

Radius (cm) =____________

This image is 120 arcseconds in height (the box of the image is L image

). A

ratio of the radius of the ring, Rring, and the image size, Limage, is equal to a ratio of

the ring’s angular size, ∝, to the angular size of the image (120 arcseconds). In symbols:

∝ = 𝑅𝑟𝑖𝑛𝑔[𝑐𝑚]

𝐿𝑖𝑚𝑎𝑔𝑒 [𝑐𝑚] × 120[𝑎𝑟𝑐𝑠𝑒𝑐]

3. Input the values you recorded in Question (2) into the equation above to find

the angular size of the nebula, ∝. Show your work, and record ∝ here:

Show your work:

4. Now apply the small angle formula, shown below, to find the actual radius (R)

of the nebula in kilometers. The distance (d) from Earth to M57 is about 2000

light years (convert this value into kilometers). The small angle formula is

𝑅 = 𝑑 × ∝

206 265

Show your work:

5. We can assume that the nebula has been expanding at about 20 kilometers

per second (km/s) since it first began losing mass. Given this speed and the size

(radius) of the nebula (which you derived in Question 4) find the time of this

expansion of the nebula (in other words, the age of M57) in seconds. Then,

convert this time (age) into years.

Show your work:

6. To better grasp the size of this object, convert the diameter of the nebula

from kilometers to astronomical units (AU), where 1 AU = 1.5 x 10 8

km. For a

sense of scale, 1 AU is the distance from the Earth to the Sun and our entire

solar system is about 200 AU across.

Show your work:

Now, explain in words these results:

PART 2: THE SIZE AND NATURE OF NGC7293, THE HELIX NEBULA

In Hubble image of NGC7293, the brighter regions are places where the

density of the nebula is high because there is more material present to create

emission. The negative image of NGC7293 is provided at the end of this

document

7. Does the nebula look like it is evenly filled, or does it look like a bubble? How

can you tell?

8. Use a ruler to measure the size of the nebula in centimeters. Measure the size

of the nebula along its longest axis. Divide this number by two to get the radius

of the nebula. Record the length of the longest axis and radius here:

Length (cm) = ___________

Radius (cm) =____________

This image is 1200 arc-seconds in height. A ratio of the radius of the ring,

Rring, and the image size, Limage, is equal to a ratio of the ring’s angular size, ∝, to the angular size of the image (1200 arc-seconds). Note the measurements here

are centimeters [cm]

∝ = 𝑅𝑟𝑖𝑛𝑔[𝑐𝑚]

𝐿𝑖𝑚𝑎𝑔𝑒 [𝑐𝑚] × 1200[𝑎𝑟𝑐𝑠𝑒𝑐]

9. Input the values you recorded in Question (8) into the ratio above to find the

angular size of the nebula, ∝. Show your work and record ∝ here:

Show your work:

10. Now apply the small angle formula, to find the actual radius (R) of the

nebula in kilometers. The distance (d) from Earth to NGC7293 is about 650 light

years (convert this values into kilometers),

𝑅 = 𝑑 × ∝

206 265

Show your work:

11. We can assume that the nebula has been expanding at about 20 kilometers

per second since it first began losing mass. Given this speed and the radius (R)

of the nebula, find the age of NGC7293 in seconds. Convert this age into years.

Show your work:

12. To better grasp the size of this object, convert the diameter of the nebula

from kilometers to astronomical units (AU), where 1 AU = 1.5 x 10 8

km. For a

sense of scale, 1 AU is the distance from the Earth to the Sun and our entire

solar system is about 200 AU across. Explain your results in words

Show your work:

PART 3: MASS LOSS

The volume of M57 can be calculated by finding the volume of a sphere

that would surround the entire nebula (see step 4) and then subtracting from it

the volume of a hollow sphere that would fit inside the nebula. To find the

Radius of the inner sphere, repeat similar steps 8 to 10 for the “reddish” circle

in figure at the end of this doc for M57.

Show your work:

13. Calculate the volume of M57 (Vring) by finding the volume of the overall

sphere and subtracting the volume of the hollow inner sphere using the

following equation [Note: be careful to use R values in kilometers]

𝑉𝑟𝑖𝑛𝑔 = 4

3 𝜋 [ 𝑅𝑜𝑢𝑡𝑒𝑟

3 − 𝑅𝑖𝑛𝑛𝑒𝑟 3 ]

Show your work:

14. The density of M57 is very low, 1.7 × 10−10 𝑘𝑔

𝑘𝑚3 . Multiply this density by the

volume in 13 to derive the mass of M57 in kilograms.

Show your work:

15. Convert M57’s mass into units of solar masses (1 𝑀𝑠𝑢𝑛 = 2 × 10 30 𝑘𝑔). If the

original star had been the same size as our Sun, what fraction of its mass did it

eject into the interstellar media (ISM) when it became a nebula?

Show your work:

17. The entire mass of this planetary nebula will become part of the interstellar

medium. If there are about 700 planetary nebulae in the Milky Way galaxy

estimate the mass returned to the interstellar medium each year by planetary

nebulae in the Milky Way by multiplying the mass of M57 (found in 15) by the

total number of nebulae in the Milky Way and then dividing by the lifetime of

M57 (in years) as found in step 5. Record the amount of mass returned to the

interstellar medium here. Is this a large amount of material?

Show your work:

18. New stars form at a rate of roughly 1 solar mass per year. Can planetary

nebulae be solely responsible for producing the gas from which these new stars

form? Explain why yes or why no?

NOTE: Some answers can be found here:

http://hubblesite.org/newscenter/archive/releases/1999/01/text/

http://hubblesite.org/newscenter/archive/releases/2004/32/text/