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A huge cannon is assembled on an airless planet having insignificant axial spin. The planet has a radius of 5.00 × 106 m and a mass of 1.46 × 1023 kg. The cannon fires a projectile straight up at 2000 m/s. An observation satellite orbits the planet at a height of 1000 km. What is the projectile's speed as it passes the satellite? (G= 6.67 × 10-11 N · m2/kg2)

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Answer:

1830 m /s

Step-by-step explanation:

The potential energy of the projectile at the surface of planet

- G x 1.46 x 10²³ m / 5 x 10⁶

Kinetic energy at surface

1/2 m v²

Total energy = - G x 1.46 x 10²³ m / 5 x 10⁶ + 1/2 m v²

- 6.67 X 10⁻¹¹ x 1.46 x 10²³ /5 x 10⁶ m + 1/2 m x 2000²

- 1.947 x 10⁶ m + 2m x 10⁶

= .053 x 10⁶ m

Potential energy of projectile at 1000 km height

= - G x 1.46 x 10²³ m / 6 x 10⁶

= - 1.6225 x 10⁶ m J

Total energy

= - 1.6225 x 10⁶ + 1/2 m V ²

Applying conservation of energy in gravitational field at surface and height

- 1.6225 x 10⁶ m + 1/2 m V ² = .053 x 10⁶ m

1/2 V ² = ( .053 + 1.6225 ) x 10⁶

1/2 V² = 1.6755 x 10⁶ m /s

V = 1.83 X 10³ m/s

1830 m /s

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User Yorb
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