Question

A comet of mass 2 × 10^8 kg is pulled toward the star. If the comet's initial velocity is very small, and the comet starts moving toward the star from 700,000,000 km away, how fast is it going right before it hits the surface of the star? (Assume that it does not lose any mass by melting as it approaches the star.)

Answer 1

The speed of the comet at the surface of the star is approximately 1,208,694.7 m/s

Step-by-step explanation:

Question parameter obtained online; The mass of the star, M = 5 × 10³¹ kg

Explanation;

The given mass of the comet, m = 2 × 10⁸ kg

The initial velocity of the comet, v → 0

The distance of the comet from the star, d = 700,000,000 km

The gravitational potential at d = G·M·m/d

The kinetic energy of the comet, K.E. = m·v²/2

The kinetic energy of the comet at d = m·(0)²/2 = 0

The gravitational potential at the surface of the star, R = G·M·m/R

The kinetic energy of the comet at the surface of the star, R = m·(v)²/2 = 0

Where;

M = The mass of the star = 5 × 10³¹ kg

`M_(Sun)` = The mass of the Sun = 1.989 × 10³⁰ kg

M/
`M_(Sun)` = 5 × 10³¹/(1.989 × 10³⁰) ≈ 25

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

R = The radius of the star

Therefore, we have;

m·(0)²/2 - G·M·m/d = m·v²/2 - G·M·m/R

∴ v = √((G·M·m/R - G·M·m/d)×2/m) = √(2·G·M(1/R - 1/d))

Therefore; v = (2 × 6.67430 × 10⁻¹¹ × 5 × 10³¹ × (1/R - 1/700,000,000,000))

v = 81696389149.1×√(1/R - 1/700,000,000,000).

The speed of the comet at the surface of the star, v = 81696389149.1×√(1/R - 1/700,000,000,000)

The mass radius relationship is given as follows;

`(R)/(R_(Sun)) = 1.30 * \left((M)/(M_(Sun)) \right)^{(1)/(2) }`

`R = R_(Sun) * 1.30 * \left((M)/(M_(Sun)) \right)^{(1)/(2) }`

The radius of the Sun = 696,340,000 M

∴ R ≈ 696,340,000 × 1.3 × √(25.14) = 4538865694.76

R = 4538865694.76 m

v = 81696389149.1×√(1/4538865694.76 - 1/700,000,000,000) ≈ 1208694.7 m/s