Question

The period of the earth around the sun is
1 year and its distance is 150 million km from
the sun. An asteroid in a circular orbit around
the sun is at a distance 185 million km from
the sun.
Period of the asteroid is 1.369 years.

What is the orbital velocity of the asteroid?
Assume there are 365 days in one year.
Answer in units of m/s

Answer 1

the reasonable way to solve this is to first write
185 million km in scientific notation

can you do that ?

Answer 2

well I've done 1.85*10^11

Answer 3

1.85 * 10^11 meters
always keep track of the units

that is the radius of a circle (the center is the sun)
the total distance around is the circumference
C= 2 pi r

so the distance around the sun is
C= 2 * 3.14 * 1.85 * 10^11 meters

can you figure that out, and put the answer in scientific notation?

Answer 4

Maybe but I used the \[V=2\pi r/T \] not too long ago and got an answer of 26924.1m/s

Answer 5

yes, that is the right idea. but the first step is 2pi r

what did you get for 2 pi r ?

Answer 6

the first step in simplifying
2 * 3.14 * 1.85 * 10^11 meters

is to multiply the numbers together:
2* pi *1.85 = ?

Answer 7

you should get 2*3.14*1.85= 11.618 or 11.62 rounded

now multiply by 10^11 meters

11.62*10^11 meters is the circumference

we could move the decimal point of 11.62 1 to the left to get 1.162
and increase the exponent of 10^11 to 10^12 to compensate

C = 1.162 * 10^12 meters

Answer 8

speed = distance / time

so speed = 1.162*10^12 meters/ 1 year

(sort of like furlongs per fortnight.. not a generally used way to write speed)

you want to change 1 year to seconds.
How many seconds in a year ?

Answer 9

31536000s

Answer 10

we work the problem we get
60*60*24*365= 31,536,000 sec/year

can you write that in scientific notation ?

Answer 11

3.15*10^7

Answer 12

ok so now we have
\[ \frac{ 11.62\cdot 10^{11} \text { meters} }{3.15 \cdot 10^7 \text{ sec}}\]

now simplify. 11.62/3.15 = ?

10^11 / 10^7 = ?

Answer 13

10^11 / 10^7 = 10^{11-7} we subtract exponents when dividing
= 10^4

Answer 14

36888.9

Answer 15

yes, but roughly that is 3.69 * 10^4 m/sec

Answer 16

indeed