Question

How do I solve this problem? Hint: To find the asteroid's acceleration, speed and period, use the following equations:a = G M / r2v = √(a r)T = 2πr/vBe sure to use the distance between the asteroid and the sun for the radius! a= gravitational acceleration [m/s2]G=gravitational constant [always 6.67 x 10-11 m3/kg s2]M= larger mass [kg]r =distance between masses or orbital radius [m]v=orbital velocity or speed [m/s]T=period / time required to complete one revolution [s]The period can be converted to years by using the conversion factor: 1 year = 3.154x107s.

Answer 1

First let's fill the things we know from the table:

G=6.67e-11 m3/kgs2

M=1.99e30 kg

r=7.33e11 m

Now, to find the acceleration of the asteroid we use:

`a=(GM)/(r^2)`

Plugging the values we know we have that the acceleration of the asteroid is:

`\begin{gathered} a=((6.67*10^(-11))(1.99*10^(30)))/((7.33*10^(11))^2) \\ a=2.47*10^(-4) \end{gathered}`

Therefore the acceleration is 2.47e-4 m/s2.

The velocity is given by:

`v=\sqrt[]{ar}`

then we have:

`\begin{gathered} v=\sqrt[]{(2.47*10^(-4))(7.33*10^(11))} \\ v=1.35*10^4 \end{gathered}`

Therefore the velocity is 1.35e4 m/s

Finally the period is given as:

`T=(2\pi r)/(v)`

then:

`\begin{gathered} T=(2\pi(7.33*10^(11)))/((1.35*10^4)) \\ T=3.41*10^8 \end{gathered}`

Therefore the period is 3.41e8 seconds. Now, this is the same as 10.8 years; to get this we divide the seconds by 3.154e7