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# Gravitational Potential watch

1. A rocket has a mass 1000 kg and the planet has a surface potential equal to -100 MJ kg^-1 .

i. Assume the fuel is used quickly to boost the rocket to high speed, calculate the Gravitational potential energy (GPE) of the rocket that must be increased for it to escape completely.

I got the answer to this, 10^11 J.

ii. If the rocket is only given 40000 MJ of kinetic energy from the fuel, determine gravitational potential at the point.

I also got the answer for this, its -60 MJ kg^-1.

iii. Determine the maximum distance from the surface it can reach, given the planet's radius is 2 x 10^6 m.

Alright I can't go further here, can somebody help me? hehe thx!
2. use GPE = mgh. You know what the GPE is at the surface, it's just -10^11J. And that's at a height of 2x10^6m. So then use that to determine mg, and then use that value of mg to find your new value of h.

Specifically: GPE(h1) = mgh1, GPE(h2) = mgh2....... h2 = GPE(h2)/mg = GPE(h2)/(GPE(h1)/h1) = h1 * GPE(h2)/GPE(h1) = 2x10^6m * -60/-100 = 2x10^6 * 6/10 = 3.3x10^4m by my reckoning. Although I just did all that in my head so I made have made a mistake
3. Willa is wrong (sorry Willa)

you cannot use mgh as the value of g decreases as you move further from the planet.

when all the KE has been transferred into GPE the rocket will have potential energy equal to mGM / r where r is the distance from the planet.

The potential GM / r at this point = - 60 MJ / kg

What we need of course is the mass of the planet. Knowing the radius of the planet is 2 x 10^6 m and knowing the potential at its surface is -100MJ/kg we can calculate the mass of the planet. ( I get 3 x 10^24kg)

so now r can be calculated. I get 3.33 x 10^6m, which is 1.33 x 10^6m above its surface.
4. (Original post by Drummy)
Willa is wrong (sorry Willa)

you cannot use mgh as the value of g decreases as you move further from the planet.

when all the KE has been transferred into GPE the rocket will have potential energy equal to mGM / r where r is the distance from the planet.

The potential GM / r at this point = - 60 MJ / kg

What we need of course is the mass of the planet. Knowing the radius of the planet is 2 x 10^6 m and knowing the potential at its surface is -100MJ/kg we can calculate the mass of the planet. ( I get 3 x 10^24kg)

so now r can be calculated. I get 3.33 x 10^6m, which is 1.33 x 10^6m above its surface.
oops. yes you win! But then all you do is flip the height dependances from my answer, since the equation is ~1/r rather than r. which gives 20/6 * 10^6m, which gives 3.33x10^6m as you said - and then yes you should subtract hte radius of the planet as well!

well done. much better than my answer.

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Updated: July 11, 2007
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