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For this problem, we assume that we are on planet-i. the radius of this planet is r =4200 km, the gravitational acceleration at the surface is gi =9.1 m/s 2 , and the gravitational constant g = 6.67 × x10−11 n m2 /kg2 in si units. the mass of planet-i is not given. not all the quantities given here will be used. suppose a cannon ball of mass m = 5130 kg is projected vertically upward from the surface of this planet. it rises to a maximum height h =12306 km above the surface of the planet. caution: here the gravitational acceleration decreases as the cannon ball travels away from planet-i. r h determine the kinetic energy (in joules) of the cannon ball immediately after it is fired off. answer in units of j.

Respuesta :

Demiurgos
The expression commonly used for potential gravitational energy is just simplification. It is actually just the first term in Taylor expansion of the real expression. 
In general, the potential energy of gravitational field is defined as:
[tex]U=-G \frac{mM}{r}[/tex]
Where G is universal gravitational constant, and r is the distance between the objects centers of mass. Negative sign represents the bound state.
Since we are not given the mass of the planet we have to calculate it.
[tex]F_g=G\frac{mM}{r_p^2}\\ mg=G\frac{mM}{r_p^2}\\ g=G\frac{M}{r_p^2}[/tex]
This formula can be used for any planet. It gives you the gravitational acceleration on the planet's surface. We can use it to calculate the planet's mass:
[tex]g=G\frac{M}{r_p^2}\\ M=\frac{gr_p^2}{G}=2.41\cdot 10^{24}kg[/tex]
Now we can calculate the potential energy of that cannonball when it reaches its maximum height.
[tex]U=-G \frac{mM}{r}\\ U=-G \frac{mM}{r_p+h}[/tex]
When we plug in the numbers we get:
[tex]U=-4.99\cdot 10^{10} J[/tex]
The potential energy has to be equal to the kinetic energy.
[tex]E_k=4.99\cdot 10^{10} J[/tex]