Can the gravitational force between supertankers affect their steering?

[Awesmund]

Hi, I am a Norwegian student and have a practical question. I am sorry about my English. My iPad keeps autocorrecting every word I am writing to Norwegian.

I have heard that supertankers steering besides each other has to adjust their direction because of the gravitational forces working between them. Is this right?

Wikipedia says that the largest oil tankers has a mass of 550 000 tonnes. If two of those are floating with one meter between them, the force working between them would be 6.67*10^(-11) * ( (5.5*10^8)^2) / 1^2 due to Newtons law of gravity.

That gives a force of 2.0*10^7 N.

Is that a force large enough to affect the tankers in any degree?

Thanks in advance

 

[technician]

Without checking your calculation I do know that ships sailing alongside each other are pulled together because of the faster flow of water between the ships..which causes a decrease in pressure ( force)...same principle as lift due to decrease in pressure as a result of increased speed.
In your calculation I think you have the separation as 1 metre. This does not seem realistic, the distance in the equation is the distance between the centres of mass.
I will check your calculation in detail later.

 

[Dale]

Wikipedia says that the largest oil tankers has a mass of 550 000 tonnes. If two of those are floating with one meter between them, the force working between them would be 6.67*10^(-11) * ( (5.5*10^8)^2) / 1^2 due to Newtons law of gravity.

That gives a force of 2.0*10^7 N.

The width of a ULCC is listed as 63 m ( http://maritime-connector.com/wiki/vlcc/ ), so a 1 m separation would give you r = 32.5 m at a minimum. Assuming that you can approximate them as point masses at their respective center of gravity then the maximum force would be 19 kN, which is less than the weight of 2 tonnes or 3.5 ppm. I doubt that they would even notice that.

Furthermore, ships displace the same mass of water, which would reduce the gravitational pull by the same amount. Because of that, I don't see how gravity could ever be a significant interaction between two ships, regardless of their mass.

 

[AlephZero]

The relevant distance is between the center of mass of the two ships.

According to Wikipedia the width (beam) of a T1 class ULCC is 68 meters, so your "1 meter separation" is really 69 meters. That gives a gravitational attraction force of about 4200 N = about 0.43 tonnes force on a ship with a mass of 500,000 tonnes, which would give an acceleration of about 10^-6g.

As technician said, the cause of this effect is the water flow pattern around ships that are sailing close to each other, not gravitational attraction. You can use Berniouilli's equation to estimate the force. (And it will be MUCH bigger than the gravitational attraction).

 

[Awesmund]

You are right. Calculating with the other extreme, that the masses are placed in the core of each boat, and in the core only, the calculation look like this:

Assuming each boat is 68.8 meters wide (guinnes world records) and that the distance between the boats is 1 meter.

6.67*10^(-11) * ( (5.5*10^8)^2) / 69.8^2 = 4141 N

I assume the actual force may be something in between the two extremes, maybe about (2.0*10^7 + 4141)N / 2 = 1.0*10^7 N ?

 

[SteamKing]

These tankers are physically quite large objects. The OP's calculation, which assumed an r of 1 m, is unreasonable, since the center of mass of the tanker will lie very close to the center of the vessel. Therefore, the attractive force between the two vessels must be calculated using an r equal to at least the width of the vessel.

 

[mfb]

You are right. Calculating with the other extreme, that the masses are placed in the core of each boat, and in the core only, the calculation look like this:

Assuming each boat is 68.8 meters wide (guinnes world records) and that the distance between the boats is 1 meter.

6.67*10^(-11) * ( (5.5*10^8)^2) / 69.8^2 = 4141 N

I assume the actual force may be something in between the two extremes, maybe about (2.0*10^7 + 4141)N / 2 = 1.0*10^7 N ?

You get a better approximation of "something in between" if you average the distances, and not the forces. The average force would correspond to point-masses with a separation of less than 2 meters.
In addition, those ships are quite long, so most parts are separated by much more than 60 meters. The total force is even below those 4000N.

 

[Awesmund]

I see, thank you very much.

Are there any great masses which are large or close enough to be affecting each other on earth?

 

[mfb]

Not in such a way that you would notice it without special equipment. It is possible to measure the gravitational attraction of two ~1kg-objects, or the deflection of gravitational attraction close to a mountain, but all those measurements require a very good sensitivity.

 

[vlad77]

Awesmund is correct: the two ships will hardly attract each other because of the influence of the displaced water -- which will act as a repulsion. Another reader was also on the right track, but then decided the displaced water mass of the second ship would cancel the first. But this is incorrect. What causes the first displaced water mass (of ship 1) to act as a repulsion is that it is no longer balanced by the analogous mass on the other side of ship 2 ... hence the effective repulsion.
What first got me interested in this problem is an example in a book by Ryabov. He gives the example of two ships attracting to show the feebleness of gravitation ... but he does not account for the displaced water.

 

[mfb]

We get an attraction above sea level and a net repulsion below.
This thread is 2.5 years old.