The asteroid 243 Ida has a mass of about 4.0 × 1016 kg and an average radius of

Question

The asteroid 243 Ida has a mass of about 4.0 × 1016 kg and an average radius of about 16 km (it's not spherical, but you can assume it is) Part A Calculate the acceleration of gravity on 243 Ida. Express your answer to two significant figures and include the appropriate units. PValue Units Part B What would an astronaut whose earth weight is 700 N weigh on 243 Ida? Express your answer to two significant figures and include the appropriate units. W. Value Units Part C If you dropped a rock from a height of 1.3 m on 243 Ida, how long would it take to reach the ground? Express your answer to two significant figures and include the appropriate units. ? Value Units Request Answer Part D If you can jump 62 cm straight up on earth, how high could you jump on 243 Ida? (Assume the asteroid's gravity doesn't weaken significantly over the distance of your jump.) Express your answer to two significant figures and include the appropriate units. hValue Units

Explanation / Answer

a.

a = GM/r² = (6.674 × 10¹¹ m³/(kg·s²))(4.0 × 10¹ kg) / (1.6 × 10 m)² = 0.0104 m/s² = 0.010 m/s2(2sf)

b.

Multiply the Earth weight by the ratio of Ida's gravity to Earth's gravity:

(700 N)(0.0104 m/s²)/(9.81 m/s²) = 0.742 N. = 0.74N (2sf)

c.

x = x + vt + ½at².

Here, x = 0 m (at the ground); x = 1.3 m; v = 0 m/s; a = -0.0104 m/s². Substitute:

(-0.0052 m/s²)t² + 0 + 1.3 m = 0, so

-(0.0052 m/s²)t² = -1.3 m, so

t² = 250 s², so

t = 15.81 s. = 16s (2sf)

d.

The potential energy at the top of the jump on Earth is mgh = m(9.81 m/s²)(0.62 m) = 6.0822m J.

Using 0.0104 for g on Ida, divide the energy by mg: (6.0822m J) / ((m)(0.0104 m/s²)) = h = 584.8 m. = 580m (2sf)

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