A projectile of mass 64 is shot from the surface of Planet Z with mass 4.34 x 1024 kg and radius 5.91 x 106 m by means of a very powerful cannon. If the projectile reaches a height of 3.08 x 105 m above Planet Z's surface, what was the speed of the projectile when it left the cannon?

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aristocles aristocles · Answer 1 · 2019-08-14T01:18:53+02:00

Answer:

[tex]v = 2.2 \times 10^3 m/s[/tex]

Explanation:

Gravity on the surface of the planet is given as

[tex]g = \frac{GM}{R^2}[/tex]

here we know that

[tex]M = 4.34 \times 10^{24} kg[/tex]

[tex]R = 5.91 \times 10^6 m[/tex]

now we have

[tex]g = \frac{(6.67 \times 10^{-11})(4.34 \times 10^{24})}{(5.91 \times 10^6)^2}[/tex]

[tex]g = 8.3 m/s^2[/tex]

now by energy conservation we have

initial KE + initial gravitational PE = final gravitational PE

[tex]\frac{1}{2}mv^2 - \frac{GMm}{R} = - \frac{GMm}{R+h}[/tex]

[tex]v^2 = 2(\frac{GM}{R} - \frac{GM}{R + h})[/tex]

[tex]v^2 = 2(6.67 \times 10^{-11})(4.34 \times 10^{24})(\frac{1}{5.91 \times 10^6} - \frac{1}{5.91 \times 10^6 + 3.08 \times 10^5})[/tex]

[tex]v = 2.2 \times 10^3 m/s[/tex]