Answer:
1830 m /s
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