Question

Two stars of masses 3 × 10³¹ kg each, and at distance 2 × 10¹¹ m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is : (Take Gravitational constant G = 6.67 × 10⁻¹¹ Nm² kg⁻²)
(A) 2.4 × 10⁴ m/s (B) 1.4 × 10⁵ m/s
(C) 3.8 × 10⁴ m/s (D) 2.8 × 10⁵ m/s

Answer 1

Thus the minimum speed of the meteorite is V = 2.8 x 10^5 m/s

Option (D) is correct.

Explanation:

• Masses of stars = 3 x 10^-3 Kg
• Distance between the masses = 2 x 10^11 m
• Radius of mass "r" = 1 x 10^11 m

- GmM / r + - GmM / r + 1/2 mv^2 = 0

2 GMm / r = 1/2 mv^2

V = √4 GM / r = √ 4 x 6.6 x 106-11 x 3 x 10^-3 / 1 x 10^11

V = 2.8 x 10^5 m/s

Thus the minimum speed of the meteorite is V = 2.8 x 10^5 m/s

Answer 2

Given :

- Mass of both stars, M = 3 × 10³¹ kg

- Distance between them, D = 2 × 10¹¹ m

- Distance from the center O to star = 1 × 10¹¹ m

- Gravitational constant, G = 6.67 × 10⁻¹¹ Nm² kg⁻²

To find :

• The minimum speed that meteorite should have at O

Solution :

• Escape velocity is defined as the lowest velocity which a body must have in order to escape the gravitational attraction of a particular planet or other object.

• Given by formula,

$Ve =\sqrt{2GM \div R} = \sqrt{gM\div R}$

For earth, Ve = 11.2 km/s

• By using energy conservation,

-GMm / + (-GMm / r) = 1/2×mVe²

where, m = mass of meteorite

1/2×mVe² = 2GMm / r

Ve² = 4GM/r

Ve = √4GM/r

Ve = 2√6.67 × 10⁻¹¹Nm² kg⁻² × 3 × 10³¹ kg / 1 × 10¹¹ m

• Ve = 2.8 × 10⁵ m/s