Question

The escape velocity (in meters per second) on Mars is \(\sqrt{\frac{2\left(6.67 \times 10^{-11}\right)\left(6.42 \times 10^{23}\right)}{3.37 \times 10^{6}}}\) Will an object traveling at 6000 meters per second escape the gravitational field of Mars?

Short answer

No, the object with a velocity of 6000 m/s will not escape Mars' gravitational field if the escape velocity is greater than 6000 m/s; otherwise, it will.

Step-by-step solution

Calculate Escape Velocity

Apply the given values into the escape velocity formula: \(V_e = \sqrt{\frac{2\left(6.67 \times 10^{-11}\right)\left(6.42 \times 10^{23}\right)}{3.37 \times 10^{6}}}\).

Solve for Escape Velocity

Perform the calculations to find the numerical value of \(V_e\).

Compare Velocities

Compare the calculated escape velocity (\(V_e\)) with the velocity of the object in question (6000 m/s). If the object's velocity is greater than or equal to \(V_e\), it will escape Mars' gravitational pull.

Key concepts

Gravitational Field

When you think about the gravitational field of a planet, imagine a net that surrounds the planet, pulling everything towards its center. The stronger the field, the harder it is to break free from the planet's gravitational pull. On Earth, we experience this pull as our weight, but when it comes to planets like Mars, it affects how hard a spacecraft must work to leave the surface.

The escape velocity is the speed needed to 'break free' from a planet's gravitational field without further propulsion. It's essentially the cosmic speed limit for leaving a planet without having to burn more fuel. This calculation holds significant importance in the realms of aerospace engineering and astrophysics, as it determines the necessary speed for a spacecraft—or any object—to exit a planet's gravitational influence without further acceleration.

Algebraic Calculation

Algebraic calculation is the mathematical manipulation of equations and formulas to find unknown values. In the case of escape velocity, we use a specific algebraic formula that includes the mass of the planet (\( M \)), the gravitational constant (\( G \)), and the radius of the planet (\( r \)). For Mars, this calculation takes the form: \[ V_e = \sqrt{\frac{2(GM)}{r}} \]

To solve for the escape velocity (\( V_e \)), we plug the values of these constants and the radius of Mars into this algebraic equation. The process involves squares, square roots, and understanding the order of operations—core components of algebra that are vital in physics problems.

Physics in Calculus

The merger of physics and calculus opens the door to dynamic systems and brings a deeper understanding of motion, forces, and energy. Calculus allows us to describe change and motion with precision. For instance, when we calculate the escape velocity from a gravitational field, we're solving a problem about how fast something needs to move to change from a state of being bound to a planetary body to being free in space.

Using calculus, we can derive equations of motion and solve them to understand the behavior of objects in gravitational fields. This not only includes escape velocity but also the paths of planets around the sun (orbits), the trajectory of a ball thrown into the air, and the way a satellite circles Earth. Physics uses calculus as a foundational tool to unlatch the complexities of these natural phenomena.