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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 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 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.

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(a) The initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

(b) The speed of the satellite is 50.24 m/s.

Acceleration due to gravity of the planet

g = GM/R²

where;

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

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²

g = 0.13 m/s²

Initial speed of the rock

v² = u² - 2gh

where;

  • v is final velocity
  • u is initial velocity

at maximum height, v = 0

u² = 2gh

u = √2gh

u = √(2 x 0.13 x 1,440)

u = 19.35 m/s

Speed of the satellite

v = √GM/r

  • M is mass of the planet Glob
  • r is the total distance from the center of the planet Glob

r = radius of planet Glob + radius of the satellite

r = 63200 m + 145,000 m = 208,200 m

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(208,200)]

v = 50.24 m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

The speed of the satellite is 50.24 m/s.

Learn more about speed of satellite here: https://brainly.com/question/22247460

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