# A 5 kg piece of a comet is at -50 C at a distance of 36000 km from the surface of the Earth. When it lands its temperature is 2500 C. What is the specific heat of the material that makes up the piece.

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### 1 Answer

By definition the **intensity of the gravitational field** generated by mass M is

`Gamma= G*(M)/R^2`

and the **gravitational potential** associated with mas M at distance R is

`U = Gamma*R = G*M/R`

Knowing the gravitational potential U we can write the potential energy of mass m as:

`E_p = m*U = G*(m*M)/R`

Now for the values

`G = 6.67*10^-11 m^3/(kg*s^2)`

`m = 5 kg` and `M =5.97*10^24 kg`

`R_(Earth) = 6370 km =6370*10^3 m`

`R = 36000 km =36000*10^3 m `

we have a difference in potential energy of

`E_p(R_(Earth)) - E_p(R_(Earth)+R) =G*m*M*(1/R_(Earth) -1/(R_(Earth) + R))=`

`= 6.67*10^-11*5*5.97*10^24 *(1/(6370*10^3) -1/((6370+36000)*10^3)) =`

`=2.656*10^8 J`

This energy is used to heat up the comet of mass m.

`Delta(E_p) =m*c*(Delta(T)) `

taking `Delta(T) = 2500 -(-50) =2550 C degree = 2550 K degree`

we obtain

`c = 2.656*10^8/(2550*5) =2.08*10^4 J/(kg*K)`

To compare with substances in nature the specific heat of granite, SiO2 and Al2O3 (Earth crust) is about `10^3 J/(kg*K)` .

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