Question

On July 4, 2005, the NASA spacecraft Deep Impact fired a projectile onto the surface of Comet Tempel 1. This comet is about 9.0 km across. Observations of surface debris released by the impact showed that dust with a speed as low as 1.0 m/s was able to escape the comet.
1. Assuming a spherical shape, what is the mass of this comet? ( The escape speed for an object at the surface of Earth is 11.2 km/s).

Express your answer using two significant figures.
M = ____________ kg
2. How far from the comet's center will this debris be when it has lost 60% of its initial kinetic energy at the surface?

Express your answer using two significant figures.
r = ____________ km

Answer 1

Answer:

1. M = 67,422,800,892,977.54 kg

2. r = 15 km

Explanation:

The diameter of the Comet Tempel 1, D = 9.0 km across

The speed with which the dust escapes = 1.0 m/s

1. The escape velocity, $v_e$, is given by the following formula

$v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{R} }$

Where;

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

$v_e$ = The escape velocity of the debris = 1.0 m/s

M = The mass of the comet from where the debris escapes

From the escape velocity equation, we have;

$M = \dfrac{v_e^2 \cdot R}{2 \cdot G}$

Plugging in the values for the variables, we get the mass of the comet, 'M', as follows;

$M = \dfrac{1.0^2 \times 9,000}{2 \cdot 6.67430 \times 10^{-11}} \approx 67,422,800,892,977.54 \, kg$

The mass of the comet, M ≈ 67,422,800,892,977.54 kg

2. When the debris has lost 60% of its initial kinetic energy, we have;

$60\% \, K.E. = 0.6\cdot K.E. = 0.6 \times \dfrac{1}{2} \times m \times v_e^2 = \dfrac{G \cdot M \cdot m}{r}$

$\therefore \, The \ distance \ of \ debris \ from \ the \ center, \, r = \dfrac{G \cdot M }{0.6 \times \dfrac{1}{2} \times v_e^2 }$

$r = \dfrac{6.67430 \times 10^{-11} \times 67,422,800,892,977.54}{0.6 \times \dfrac{1}{2} \times 1^2 } = 15,000$

When the debris has lost 60% of its initial kinetic energy, the distance the debris will be from the comet's center, r = 15,000 m = 15 km