# Eureka Math Grade 8 Module 1 Lesson 10 Answer Key

## Engage NY Eureka Math 8th Grade Module 1 Lesson 10 Answer Key

### Eureka Math Grade 8 Module 1 Lesson 10 Exercise Answer Key

Exercise 1.
The speed of light is 300,000,000 meters per second. The sun is approximately 1.5×1011 meters from Earth. How many seconds does it take for sunlight to reach Earth?
300 000 000=3×108
$$\frac{1.5 \times 10^{11}}{3 \times 10^{8}}$$ = $$\frac{1.5}{3}$$×$$\frac{10^{11}}{10^{8}}$$
=0.5×103
=0.5×10×102
=5×102
It takes 500 seconds for sunlight to reach Earth.

Exercise 2.
The mass of the moon is about 7.3×1022 kg. It would take approximately 26,000,000 moons to equal the mass of the sun. Determine the mass of the sun.
26 000 000=2.6×107
(2.6× 107)(7.3×1022)=(2.6×7.3)(107×1022)
=18.98×1029
=1.898×10×1029
=1.898×1030
The mass of the sun is 1.898×1030 kg.

Exercise 3.
The mass of Earth is 5.9×1024 kg. The mass of Pluto is 13,000,000,000,000,000,000,000 kg. Compared to Pluto, how much greater is Earth’s mass than Pluto’s mass?
13 000 000 000 000 000 000 000=1.3×1022
5.9×1024-1.3×1022=(5.9×102)×1022-1.3×1022
=(590-1.3)×1022
=588.7×1022
=5.887×102×1022
=5.887×1024
The mass of Earth is 5.887×1024 kg greater than the mass of Pluto.

Exercise 4.
Using the information in Exercises 2 and 3, find the combined mass of the moon, Earth, and Pluto.
7.3×1022+1.3×1022+5.9×1024=(7.3×1022+1.3×1022)+5.9×1024
=8.6×1022+5.9×1024
=(8.6+590)×1022
=598.6×1022
=5.986×102×1022
=5.986×1024
The combined mass of the moon, Earth, and Pluto is 5.986×1024 kg.

Exercise 5.
How many combined moon, Earth, and Pluto masses (i.e., the answer to Exercise 4) are needed to equal the mass of the sun (i.e., the answer to Exercise 2)?
$$\frac{1.898 \times 10^{30}}{5.986 \times 10^{24}}$$ = $$\frac{1.898}{5.986}$$ × $$\frac{10^{30}}{10^{24}}$$
=0.3170…×106
≈0.32×106
=0.32×10×105
=3.2×105
It would take 3.2×105 combined masses of the moon, Earth, and Pluto to equal the mass of the sun.

### Eureka Math Grade 8 Module 1 Lesson 10 Problem Set Answer Key

Have students practice operations with numbers written in scientific notation and standard notation.

Question 1.
The sun produces 3.8×1027 joules of energy per second. How much energy is produced in a year? (Note: a year is approximately 31,000,000 seconds).
31 000 000=3.1×107
(3.8×1027)(3.1×107)=(3.8×3.1)(1027×107)
=11.78×1034
=1.178×10×1034
=1.178×1035
The sun produces 1.178×1035 joules of energy in a year.

Question 2.
On average, Mercury is about 57,000,000 km from the sun, whereas Neptune is about 4.5×109 km from the sun. What is the difference between Mercury’s and Neptune’s distances from the sun?
57 000 000=5.7×107
4.5×109-5.7×107=(4.5×102)×107-5.7×107
=450×107-5.7×107
=(450-5.7)×107
=444.3×107
=4.443×102×107
=4.443×109
The difference in the distance of Mercury and Neptune from the sun is 4.443×109 km.

Question 3.
The mass of Earth is approximately 5.9×1024 kg, and the mass of Venus is approximately 4.9×1024 kg.
a. Find their combined mass.
5.9×1024+4.9×1024=(5.9+4.9)×1024
=10.8×1024
=1.08×10×1024
=1.08×1025
The combined mass of Earth and Venus is 1.08×1025 kg.

b. Given that the mass of the sun is approximately 1.9×1030 kg, how many Venuses and Earths would it take to equal the mass of the sun?
$$\frac{1.9 \times 10^{30}}{1.08 \times 10^{25}}$$=$$\frac{1.9}{1.08} \times \frac{10^{30}}{10^{25}}$$
=1.75925…×105
≈1.8×105
It would take approximately 1.8×105 Venuses and Earths to equal the mass of the sun.

### Eureka Math Grade 8 Module 1 Lesson 10 Exit Ticket Answer Key

Question 1.
The speed of light is 3×108 meters per second. The sun is approximately 230,000,000,000 meters from Mars. How many seconds does it take for sunlight to reach Mars?
230 000 000 000=2.3×1011
$$\frac{2.3 \times 10^{11}}{3 \times 10^{8}}$$=$$\frac{2.3}{3}$$×$$\frac{10^{11}}{10^{8}}$$
=0.7666…×103
≈0.77×10×102
≈7.7×102
It takes approximately 770 seconds for sunlight to reach Mars.

Question 2.
If the sun is approximately 1.5×1011 meters from Earth, what is the approximate distance from Earth to Mars?