From 1-3: Mass of Asteroid Eros you learned that its mass can be determined from Kepler s Third Law. Here are the results from the data collected on three different days: Date Mass of Eros (kg) February 14, 2000 6.75 x 10 15 March 3, 2000 6.83 x 10 15 April 11, 2000 6.71 x 10 15 What is the average mass of Eros? Question 1 options: 6.76 kg 6.76 x 10 15 kg 20.29 x 10 15 kg Question 2 Eros is 33 km long, and 13 km across. If you use these dimensions and the formula for the volume of a perfect cylinder, (V = pi x r 2 x h), you find that the volume is 4.4 x 10 12 m 3 . The NEAR scientists, however, estimate the volume to be 2.6 x 10 12 m 3 . Why are the two values different? (Hint: Draw a shape of cylinder our calculation is based on and then compare it with the shape of Eros in the photo. What difference is there?) Question 3 (2 points) Since Density = mass / volume, use the average mass from question#1 above and the volume estimated by the NEAR scientiest in question #2 above to detremine the average density of Eros. (Note: keep in mind our result here is the average density. Eros could have denser materials in its core than its surface.) 2.6 x 10 27 kg/m 3 2600 kg/m 3 1.06 kg/m 3 Solution 1) m avg = m1+m2+m3/3 m avg = 6.76*10^15 kg 2) the two values are different because the first estimate the volume for a perfect cylinder. Eros is not a perfect cylinder it lokks like if it is rounded off at the ends . so two values are different 3) density = m/v d = 6.76*10^15/2.6*10^12 = 2600 kg/m^3 .

1. From 1-3: Mass of Asteroid Eros you learned that its mass can be determined from Kepler s
Third Law. Here are the results from the data collected on three different days:
Date
Mass of
Eros
(kg)
February 14, 2000
6.75 x 10 15
March 3, 2000
6.83 x 10 15
April 11, 2000
6.71 x 10 15
What is the average mass of Eros?
Question 1 options:
6.76 kg
6.76 x 10 15
kg
20.29 x 10 15
kg
Question 2
Eros is 33 km long, and 13 km across. If you use these dimensions and the formula for the
volume of a perfect cylinder, (V = pi x r 2
x h), you find that the volume is 4.4 x 10 12
m 3
. The
NEAR scientists, however, estimate the volume to be 2.6 x 10 12
m 3
. Why are the two values
different? (Hint: Draw a shape of cylinder our calculation is based on and then compare it with
the shape of Eros in the photo. What difference is there?)
Question 3 (2 points)
Since Density = mass / volume, use the average mass from question#1 above and the volume
estimated by the NEAR scientiest in question #2 above to detremine the average density of Eros.
(Note: keep in mind our result here is the average density. Eros could have denser materials in its
core than its surface.)

2. 2.6 x 10 27
kg/m 3
2600 kg/m 3
1.06 kg/m 3
Date
Mass of
Eros
(kg)
February 14, 2000 6.75 x 10 15
March 3, 2000 6.83 x 10 15
April 11, 2000 6.71 x 10 15
Solution
1) m avg = m1+m2+m3/3
m avg = 6.76*10^15 kg
2) the two values are different because the first estimate the volume for a perfect cylinder. Eros is
not a perfect cylinder it lokks like if it is rounded off at the ends . so two values are different
3) density = m/v
d = 6.76*10^15/2.6*10^12 = 2600 kg/m^3