How many drops in all the oceans?

  • Thread starter Thread starter Lemniscate
  • Start date Start date
AI Thread Summary
The discussion centers on calculating the number of water drops in the Earth's oceans, with the assumption that 1 cm³ contains 25 drops. The average ocean depth is 4 km, and the Earth's radius is 6400 km, with oceans covering about 70% of the surface. The user initially struggles with whether to use the surface area or volume of a sphere for their calculations. After some confusion regarding unit conversions, they realize that 1 km³ equals 10^15 cm³, not 10^5 cm³, which impacts their final calculations. The user acknowledges their mistake and expresses gratitude for the assistance received.
Lemniscate
Messages
16
Reaction score
0

Homework Statement



Simple question, but I don't know how to correctly approach it.

How many drops of water are in all the oceans on earth? Assume that 1cm^3 contains 25 drops of water.


Homework Equations



The average depth of the ocean is 4 km. I know I will have to convert to cm.
Radius of Earth: 6400 km
About 70% of the Earth is covered by oceans.


The Attempt at a Solution



Need help setting up the problem correctly.
I've tried various things. I'm not sure if I have to use the surface area of a sphere or the volume of a sphere.

I have to subtract the depth of the oceans (4km) from 6400km...?
 
Physics news on Phys.org
Find the surface area of the earth, then take the proportion of this that is covered by water.
Then imagine unrolling this areas as a flat map, not find the volume of a slab of water of this flat area and 4km high.

Then convert km^3 to cm^3
 
mgb_phys,
by doing what you advised, I obtained a surface area of 5.15 x 10^8 km^2 times .70, which gave me a total of 3.6 x 10^8 km^2 roughly.

3.6 x 10^8 km^2 times 4 km for the depth gives me a volume of 1.44 x 10^14 cm ^3 (once I conver to cm).

Am I missing something? Because this is not the correct answer.

The question is asking me for the amount of drops of water. I should be multiplying my final answer, 1.44 x 10^14 cm ^3, by 25 (1cm^3 = 25 drops).

Where am I making a mistake?
 
1 km^3 = (1 00 000 cm)*(1 00 000 cm)*(1 00 000 cm)=10^15 cm^3, not 10^5 cm^3.
 
Oops. I see what I did wrong.

Thanks to both of you!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Replies
6
Views
2K
Replies
5
Views
3K
Replies
3
Views
14K
Replies
13
Views
2K
Replies
4
Views
3K
Replies
3
Views
5K
Replies
11
Views
2K
Back
Top