Question

Circular motion: check my work please?

Answer 1

Okay, the first question asks the orbital speed is of a satellite orbiting 300 km away (3.0 x 10^5 m). Information given: massearth = 5.98 x 10^24 kg. Use the formula: v = sqrt of (G)(m)/r. Plug in, get: v = 3.65 x 10^4 m/s.

It then asks you how long it would take to complete one orbit, I am unsure how to find that.

Answer 2

Second question asks the orbital speed of Neptune around the Sun, and how long it would take to complete one orbit. Used the same formula as above, with the following information: mass-sun = 1.99 x 10^30 kg, radius = 4.50 x 10^12 km (4.50 x 10^15 meters). Plugged into formula and got 1.71 x 10^2 m/s.

Again, unsure of how to find the time it would take to complete one orbit, unless you use a kinematic equation?

Answer 3

Final question asks the orbital speed and time of one complete orbit. I used the same equation above with the following information: mass-earth = 5.98 x 10^24 kg, radius = 5.0 x 10^4 m. Plugged into the formula and got: 8.93 x 10^4 m/s.

Still unsure how to find one complete orbit time, lol.

Answer 4

Oops, forgot to mention for the final question: the satellite is 50 km away from the surface of the earth (in my original comment I just converted it to meters).

Answer 5

I didn't check your speeds yet, but let's assume you managed to plug the numbers in correctly.

As you wrote at the very beginning, you're assuming circular motion. Meaning the orbiting object will describe a circle with radius r around a point (center of earth or sun).
Since you know that speed is distance over time, you can calculate the time for one orbit by finding the distance the object has to travel (->once around the circle<-) and dividing it by the speed it's going at.

\[v = \frac{d}{t}\]
\[t = \frac{d}{v}\]

Answer 6

In the equations above, v is the speed, d is the distance and t is the time.

Answer 7

Okay, so would I use the circumference of the earth and sun for the three problems?

\[t = \frac{ 2\pi r }{ v }\]

Answer 8

You'd use the circumference of the orbit, yes. (your equation is correct)

Answer 9

The speeds you calculated are also correct, it seems :)

Answer 10

Okay, so:

1st question: 5.16 x 10^1

2nd question: 1.65 x 10^14

3rd question: 3.52

Answer 11

First off, you're missing your units.

Second: if you think about the result, something is wrong. (3.5 seconds for one orbit doesn't seem right). What is wrong is the following:

If the satellite is 50km from the surface of the earth, it is 50km + radius_of_earth from the center of the earth (around which we assume it is orbiting). The radius of the earth is ~6370 km, so we got the distances wrong.

I'm sorry I didn't catch this earlier - should've though about it more carefully. :-/

Answer 12

For the last question would I need to redo both the orbital speed and time for one orbit, or just the orbit time?

Answer 13

Since the (wrong) radius was used to calculate the orbital speed, you'll need to redo that as well. Morevover, since you're using the mass of the earth in question one as well, I'm assuming the question reads "satellite orbiting earth", too?

Answer 14

Okay, then:

r-earth = 6.38 x 10^6 m
r-satellite-earth = 5.0 x 10^4 m
r-total = 6.43 x 10^6

\[\sqrt{\frac{ (6.67 \times 10^{-11}) \times (5.98 \times 10^{24}) }{ 6.43 \times 10^{6} }}\]

So, \[v = 7.88 \times 10^{3}\] m/s

Then for orbit time:

\[t = \frac{ 2 \pi (6.43 \times 10^{6}) }{ 7.88 \times 10^{3} }\]

So, \[t = 5.13 \times 10^{3} s\]

Answer 15

That is correct. Depending on the exact numbers you use, it'll change, but I got:

Qestion 1:\[v_1 = 7730 \; \frac{\text{m}}{\text{s}}\]
\[t_1 = 5423 \; \text{s} = \text{1 hour 30 minutes 23 seconds}\]
Qestion 2:
\[v_2 = 7879 \; \frac{\text{m}}{\text{s}}\]
\[t_2 = 5121 \; \text{s} = \text{1 hour 25 minutes 21 seconds}\]

Note: Your numbers are perfectly fine, too - it just depends on what exactly you plug in for G and the mass and radius of the earth.

Thinking about those numbers, they are much more realistic. If we compare the result of question one with reality for example, the International Space Station is about 410km above the surface of the earth and has an orbital periode of 93 minutes (1 hour 33 min). That's very, very close to what we got for our period at 300km and much more realistic than 51.6 seconds ;)

Answer 16

Thank you so much for your help!

Answer 17

You're very welcome.

final remark:
As you can see, it's always a good idea to pause for a minute and think about your results. If you are way off from what you would've expected, you might have made a mistake.

Answer 18

Actually, did I do something wrong regarding question two? We both got an hour or so for the orbit (around the sun), but my research tells me Neptune's orbit is really closer to a century, nearly two.

Answer 19

I don't want to redo this! Lol.

Answer 20

I thought, you got 1.6*10^14 seconds? But yes, that's wrong.

The radius isn't 4.5*10^15 meters, but 4.5*10^12 meters. Orbital speed comes out to 5433 m/s and the time comes out to 5.204*10^9 seconds. That's around 164.9 years.

Answer 21

Well, then. Maybe I shouldn't be converting units, thank you very much!

Answer 22

Does your problem say 4.5*10^12 km? Because then your teacher made a mistake. Wikipedia says the radius is ~30AU (astronomical units - 1AU is the distance earth to sun). That comes out to 4.5*10^9 km.

You should always convert your units to SI units, like you did. That's a very good idea and extremely helpful. Keep doing it!

Answer 23

Yeah, I always make sure to convert my units. And the worksheet said 4.50 x 10^12 km.