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The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.

Respuesta :

onyebuchinnaji

The initial speed of the rock as it left the astronaut's hand is 168 m/s.

The speed of the satellite is 60.2 m/s.

Acceleration due to gravity of the satellite

g = GM/R²

where;

  • M is mass of the satellite
  • R is radius of the satellite

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(6.32 x 10⁴)²

g = 0.132 m/s²

initial speed of the rock when it reaches maximum height

v² = u² - 2gh

0 = u² - 2gh

u² = 2gh

u = √2gh

u = √(2 x 9.8 x 1440)

u = 168 m/s

Speed of the satellite

v = √GM/r

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(1.45 x 10⁵)]

v = 60.2  m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 168 m/s.

The speed of the satellite is 60.2 m/s.

Learn more about speed of satellite here: https://brainly.com/question/25721729

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