Escape the earth

[Nano]

Hello. Can someone help me with how to calculate how much energy (in joules) it would take to lift an object with a given mass and radius out of the earths gravitational field? I get the answer in m\s (escape velocity), but I need to know what correspond to that in energy.

 

And I also wonder how to calculate how much fuel is needed to make an object escape the earth.

 

Thanks.

[Sisyphus]

Well, it's simple enough if you disregard air resistance. Kinetic energy is equal to (1/2)mv^2, with energy in joules, mass in kilograms, and velocity in meters per second. Including air resistance makes it a lot more complicated, since it's dependent on the shape of the object.

[insane_alien]

just find out your gravitational potential energy (not the mgh stuff) and there you go. that is the minimum energy needed to remove you from the earth.

[Nano]

Thanks for answers. Is this right then? If I want to send an object of 1 kg to Mars, when the planet is the closest to the earth as possible:

 

Ug = mgh

 

1kg x

9.8^2 m\s x

54,6 x 10^9 m =

5,2*10^12 Joules

 

But this formula does not take into account that the gravitational field gets weaker the further away from the earth you go?

 

Do I also have to think about that the sun is pulling this object (If I choose to send it toward Mars AWAY from the sun)

 

What is the corresponding amount of Joules stored in 1 liter of jet-engine-fuel? And how much does 1 liter of fuel weigh?

[insane_alien]

i said not the mgh one.

 

U=-Gm1m2/r

 

that one.

[Janus]

Thanks for answers. Is this right then? If I want to send an object of 1 kg to Mars, when the planet is the closest to the earth as possible:

 

Ug = mgh

 

1kg x

9.8^2 m\s x

54,6 x 10^9 m =

5,2*10^12 Joules

 

But this formula does not take into account that the gravitational field gets weaker the further away from the earth you go?

 

Right, which is why you use the formula given by insane_alien instead.

 

Do I also have to think about that the sun is pulling this object (If I choose to send it toward Mars AWAY from the sun)

 

Yes, you do. However, to use the minimum amount of energy to get to Mars you do not launch when Mars is at its closest. Instead you use a Hohmann transfer orbit. This type of orbit intersects the orbit of Mar's 180° from where it leaves Earth.

The velocity needed to enter this orbit is found by:

 

[math] \Delta V = V_e \left ( \sqrt{\frac{2Rm}{Re+Rm}}+1 \right )[/math]

Where Ve = the orbital velocity of the Earth

Re the orbital radius of the Earth

Rm the orbital radius of Mars

 

The energy can then be found by

 

[math]E = \frac{\Delta V^2 M}{2}[/math]

 

What is the corresponding amount of Joules stored in 1 liter of

About 3.4 million Joules

jet-engine-fuel? And how much does 1 liter of fuel weigh?

787 grams.

[Nano]

Great, thanks. Lots of help.

[swaha]

what is hoghmann transfer orbit? can anybody give me the deduction for minimum energy? or tell me where to find it pls. i am a 1st yr stdnt of physics only.

[marketsmarkets]

By finding gravitational PE. We can get ans.

[swansont]

what is hoghmann transfer orbit? can anybody give me the deduction for minimum energy? or tell me where to find it pls. i am a 1st yr stdnt of physics only.

 

Google leads me to Wikipedia. These two sites are your friends for this kind of question.

 

http://en.wikipedia.org/wiki/Hohmann_transfer_orbit