Question

Problem 13.46 < 15 of 17 Review | Constants Suppose that on earth you can jump straight up a distance of 57 cm. Asteroids are made of material with mass density 2800 kg/m² Part A What is the maximum diameter of a spherical asteroid from which you could escape by jumping? Express your answer with the appropriate units. HÅ BW ? Value Units Submit Request Answer

Answer 1

Given:

• Maximum height to jump :
• Density of the asteroid :
  p=2800 kg/m
  
• Acceleration due to gravity :
  9 = 9.81 m/s
  
• Gravitational constant : $G = 6.67 \times 10^{-11} \ m ^{3} / kg \cdot s^{2}$

From this given height of jumping we can find the initial velocity. Since the final velocity will become zero, we can write,

Vi = 2gh

The density is defined as the mass per unit volume,

$\rho = \frac{M}{V}$

$M = V \rho$

$M = \frac{4}{3} \pi r ^{3} \rho$

Where, r is the radius of the asteroid.

Now, by using conservation of energy we can have,

$\frac{1}{2} m ( v_{i } ) ^{2} = \frac{GM m}{r}$

$\frac{1}{2} ( v_{i } ) ^{2} = \frac{GM }{r}$

By plugging the values of initial velocity and mass in the above expression we get,

$\frac{1}{2} ( \sqrt { 2gh } ) ^{2} = \frac{G \: \left (\frac{4}{3 } \pi r ^{3} \rho \right ) }{r}$

$r ^{2} = \frac{3 gh }{4G \pi \rho }$

$r = \sqrt { \frac{3 \times 9.81 \times 0.57 } {4 \times 6.67 \times 10^{-11} \times \pi \times 2800 } }$

$r = 2.6735 \times 10^{3}$

This is the radius of the asteroid.

Maximum diameter is given by,

$d_{max} = 2 r = 2 \times 2673.5$

$\boldsymbol{d_{max} = 5347 \ m }$

or

$\boldsymbol{d_{max} = 5.35 \ km }$

This is the maximum diameter of a spherical asteroid from which you could escape by jumping.