Question

A spaceship is idling at the fringes of our galaxy, 80,000 light-years from the galactic center. What minimum speed must it have if it is to escape entirely from the gravitational attraction of the galaxy? The mass of the galaxy is \(1.4 \times 10^{11}\) times that of our Sun. Assume, for simplicity, that the matter forming the galaxy is distributed with spherical symmetry.

Short answer

The spaceship must have a minimum speed of \(31622.776\, \text{m/s}\) or round up to \(31623 \, \text{m/s}\) to entirely escape the gravitational attraction of the galaxy.

Step-by-step solution

Convert solar mass to kilograms

One solar mass (\(M_{\odot}\)) is approximately \(2 \times 10^{30}\) kilograms. Therefore, the mass of the galaxy (M) is \(M = 1.4 \times 10^{11} M_{\odot} = 1.4 \times 10^{11} \times 2 \times 10^{30} = 2.8 \times 10^{41}\) kilograms.

Convert light-years to meters

One light-year is approximately \(9.461 \times 10^{15}\) meters. Therefore, the distance from the center of the galaxy (r) is \(r = 80,000 \, \text{light-years} = 80,000 \times 9.461 \times 10^{15} = 7.5688 \times 10^{21}\) meters.

Calculate the escape velocity

Substitute the values of G, M, and r into the escape velocity formula to find the minimum speed required to escape the galaxy \(v = \sqrt{\frac{{2 \times 6.67 \times 10^{-11} \times 2.8 \times 10^{41}}}{7.5688 \times 10^{21}}} = \sqrt{9.96 \times 10^{10}} = 31622.776\, \text{m/s}\)

Key concepts

Gravitational Attraction

Gravitational attraction is the force that pulls two masses towards each other. It's the essential force for understanding celestial movements and interactions. This fundamental force explains not only the motion of planets, moons, and stars but also how a spaceship can escape a galaxy's gravitational pull, as demonstrated in our exercise.
Newton's law of universal gravitation states that any two objects with mass attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula for gravitational force is:
\( F = G \frac{{m_1 \times m_2}}{{r^2}} \)
where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the objects, and \( r \) is the distance between the centers of the two masses. For an object to 'escape' this force from a large body like a galaxy, it needs to reach a speed where the kinetic energy is equal to the gravitational potential energy. This speed is known as the escape velocity.

Galactic Mass Calculation

Calculating the mass of a galaxy is a crucial step in determining the escape velocity needed for a spacecraft to leave its gravitational influence. The mass of an entire galaxy is immense, often measured in multiples of the mass of our Sun, known as solar masses.
In our exercise, the galaxy's mass is given as \(1.4 \times 10^{11}\times \text{Sun's mass} \) which translates into \(1.4 \times 10^{11}\times \text{Solar masses} \) when using the Sun as a unit. Converting solar masses to kilograms helps bring this number into more scientific and calculable terms.
As outlined in the solution, the mass of the galaxy is calculated using the mass of our Sun in kilograms, which allows us to apply it conveniently within Newton's law of gravitation. Galactic mass calculation is vital for understanding the dynamics within the galaxy, such as orbital mechanics, stellar evolution, and even making predictions about dark matter distribution.

Conversion of Astronomical Units

Distances in space are so vast that regular units like meters or kilometers become impractical to use. Astronomical units, light-years, and parsecs are often employed instead. A light-year is the distance that light travels in one year and is frequently used to express the gaps between stars and galaxies.
For our galactic escape problem, we convert light-years into meters to use in physical equations that determine escape velocity. This conversion is a critical step for accurate calculations involving the laws of physics, which require a consistent unit system - in this case, the SI units.
Converting 80,000 light-years into meters involves multiplying the number of light-years by the distance light travels in a year – around \(9.461 \times 10^{15} \) meters. This precise conversion provides a tangible metric distance, which can then be used in gravitational equations to provide the meaningful solution we are after, like the minimum speed needed for a spaceship to escape a galaxy's gravitational field.