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

By estimating the mass of Comet Tempel 1 using the formula for escape velocity, it is calculated to be 3.4×10¹³ kg. To determine the distance from the comet's center when debris loses 60% of its initial kinetic energy, conservation of energy principles indicate that this distance is 7.5 km.

Step-by-step explanation:

The mass of Comet Tempel 1 can be estimated by using the formula for escape velocity v = √(2GM/r), where v is the escape velocity, G is the gravitational constant (6.674×10⁻¹¹ N(m/kg)2), M is the mass of the object, and r is the radius of the object. According to the question, the escape velocity is 1.0 m/s for the lowest speed dust particles that can escape, and the comet has a radius of 4.5 km (since it's 9.0 km across). By rearranging the formula to solve for M, we get M = v2r / (2G). Plugging the values, we find M = (1.0 m/s)2 × (4.5×10³ m) / (2×6.674×10⁻¹¹ N(m/kg)2) = 3.39×10¹³ kg, which needs to be expressed with two significant figures, so M = 3.4×10¹³ kg.

For part two, the question asks at what distance from the comet's center the debris will be when it has lost 60% of its initial kinetic energy. Initially, we have the kinetic energy KE0 = 0.5 m v2, and after losing 60% of this energy, 0.4 KE0 remains. The kinetic energy at a distance r from the comet is related to the potential energy by KE + PE = constant, with the potential energy given by PE = -GMm/r. Using the conservation of energy, we can set up the equation 0.4×0.5mv2 - GMm/(4.5×10³ m) = 0.5mv2 - GMm/r, solving for r yields r = 7.5×10³ m or r = 7.5 km when rounded to two significant figures.

Answer 2

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

2. r = 15 km

Step-by-step 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{(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 = (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 = (1.0^2 * 9,000)/(2 \cdot 6.67430 * 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 * (1)/(2) * m * v_e^2 = (G \cdot M \cdot m)/(r)`

`\therefore \, The \ distance \ of \ debris \ from \ the \ center, \, r = (G \cdot M )/(0.6 * (1)/(2) * v_e^2 )`

`r = (6.67430 * 10^(-11) * 67,422,800,892,977.54)/(0.6 * (1)/(2) * 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