# Time required to go to Mars

1. Aug 2, 2004

### terpsgirl

A rocket moves through space at 11,000 m/s. At this rate, how much time will be required to go to mars (78.3 x 10^6 km)?

i used t= d/v

and got 5.05 x 10^ -4

I'm not sure what i'm doing wrong. I cant get the correct answer

2. Aug 2, 2004

### Tom Mattson

Staff Emeritus
Neither am I, because you haven't shown any of your steps.

3. Aug 2, 2004

### Staff: Mentor

Why don't you show us the exact numbers you are plugging in. Don't forget to use consistent units. For example, don't mix meters and kilometers.

4. Aug 2, 2004

### terpsgirl

11,000 m/s
__________
21,751,740 = 5.05 x 10^ -4

I got my numerator figure by converting 78.3 x 10 ^6 km
and got 21,751,740 m/s

5. Aug 2, 2004

### Tom Mattson

Staff Emeritus
You can't convert km to m/s. The former is a distance, the latter is a speed.

You should have converted 78.3 x 106 km to meters. The numerical value will only differ by a power of 10. Also, you're dividing in the wrong order. The speed v goes on the bottom, and the distance d goes on top.

6. Aug 2, 2004

### terpsgirl

I got 71181818.18 = 71 seconds as my answer. THX :)

7. Aug 2, 2004

### Tom Mattson

Staff Emeritus
The answer is t=7118181.81 seconds (which is about 82 days). You can't round that to 71 seconds (which is a minute and 11 seconds).

8. Aug 2, 2004

### BobG

Uh, I think Tom meant the they would differ by A power of 10, not that 78.3 x 10^6 km equaled 78.3 x10^10 meters. To go from km to meters, multiply by 1000 (or just change the power of ten by three - 78.3 x 10^6 to 78.3 x 10^9)

And how did 71 plus million seconds suddenly turn into only 71 seconds. You might want to convert your 7 million+ seconds into something a little more manageable such as days, for example.

9. Aug 2, 2004

### NateTG

Tracking dimensions

There's a really good technique that a physics teacher taught me (although it's probably common use.) For now, whenever you do a physics problem, you should keep track of the units.

Here's an example problem which is simlar to yours:

So, let's say we have a car that is traveling at $$5 \times 10^2 feet/sec$$ and we want to know how many days it will take to drive $$7 \times 10^5 miles$$.

So, of course, we start with$$d=vt$$. Since we want to solve for time, we divide both sides by $$v$$ so we have $$t=\frac{d}{v}$$

So, let's plug in the numbers:

$$t=\frac{d}{v}=\frac{7 \times 10^5 miles}{5 \times 10^2 feet/sec}=1.4 \times 10^3 \frac{sec \times miles}{ft}$$

Of course, second miles per foot are not appropriate units, since we want days. So we can do the following:
$$1 mile=5280 ft$$
$$1=\frac{5280 ft }{1 mile}$$
so
$$t=1.4 \times 10^3 \frac{sec \times miles}{ft} =1.4 \times 10^3 \frac{sec \times miles}{ft} \times 1 =$$
$$1.4 \times 10^3 \frac{sec \times miles}{ft} \times \frac{5280 ft}{1 mile}=7 \times 10^6 sec$$
Note that the miles and feet cancel nicely when you multiply out.
Now, to get from seconds to days, you can use a similar approach:
$$7 \times 10^6 sec \times \frac{1 minute}{60 seconds} \times \frac{1 hour}{60 minutes} \times \frac{1 day}{24 hours}\approx 1.2 \times 10^1 days$$

If you write up your problem in a similar fashion, you're less likely to make mistakes, and people here, and your teacher, will have an easier time helping you.

P.S. I was playing a bit fast and loose with sig figs there. You should be more careful.

10. Aug 2, 2004

### HallsofIvy

Staff Emeritus
Every physics teacher should teach about keeping track of units! That's a basic concept :).

By the way, I am still wodering if "NateTG" is "Nate the great" who was "12 years old" for more years than I can remember! If so he has finally grown up. His responses actually make sense now!

Last edited: Aug 2, 2004
11. Aug 3, 2004

### NateTG

Since you're in VA, probably not.