The acceleration due to gravity near a planet\'s surface is known to be 4.00m/s^

Question

The acceleration due to gravity near a planet's surface is known to be 4.00m/s^2 .

a)If the escape speed from the planet is 9.89km/s , determine its radius.

b)Find the mass of the planet.

c)If a probe is launched from its surface with a speed twice the escape speed and then coasts outward, neglecting other nearby astronomical bodies, what will be its speed when it is very far from the planet? (Neglect any atmospheric effects also.)

d)Under these launch conditions, at what distance will its speed be equal to the escape speed?

Explanation / Answer

a)

Acceleration due to gravity = 4 m/s2
4 = GMp/R2 or GMp = 4R2
escape velocity = 9890 m/s = sqrt(2GMp/R)
or 9890 = sqrt(8R2/R)
or 97812100 = 8R
or R = radius of planet = 12226512.5 m = 12226.51 km

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b) mass of planet = 4R2/G = 8.9647 * 1024kg

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b) escape velocity = 9890 m/s
Energy of launch required for escape = 0.5*mprobe* 98902
energy given = 0.5*mprobe*197802

kinetic energy of probe after escaping atmosphere= 0.5*mprobe* ( 197802 - 98902)

Let required velocity be V
0.5*mprobe*V2 = 0.5*mprobe* ( 197802 - 98902)
or V2 = 293436300
or V =speed far away = 17129.98 m/s = 17.13 km/s (ans)

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c) In abobe launch conditions ie launch velocity = twice escape velocity
we get velocity at point far away from planet = 17.13 km/s which is greater than escape speed.
Because of absence of astronomical forces, the iobject's speed will never be equal to escape velocity.

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