Question

The escape velocity for a rocket from earth is \(11.2 \mathrm{kms}^{-1}\) value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\) (A) \(11.2\) (B) \(22.4\) (C) \(5.6\) (C) \(53.6\)

Short answer

The escape velocity for a rocket from a planet where the acceleration due to gravity is double that on Earth and the diameter of the planet is twice that of Earth is \(22.4 \mathrm{kms^{-1}}\), which corresponds to option (B).

Step-by-step solution

Find the Relationship between Mass and Radius of the Other Planet

Start by setting up the gravitational force equations for both Earth and the other planet: For Earth: \(F_E = G \frac{M_E m}{R_E^2}\), For the other planet: \(F_P = G \frac{M_P m}{R_P^2}\), where \(F_E\) and \(F_P\) are the gravitational forces on Earth and the other planet respectively, \(M_E\) and \(M_P\) are their masses, and \(R_E\) and \(R_P\) are their radii. We are given that the acceleration due to gravity on the other planet is double that on Earth: \(\frac{F_P}{m} = 2 \frac{F_E}{m}\).

Solve for the Mass of the Other Planet

Cancel out m and G from both sides: \(\frac{M_P}{R_P^2} = 2 \frac{M_E}{R_E^2}\). We are given that \(R_P = 2 R_E\), so substitute it: \(\frac{M_P}{(2 R_E)^2} = 2 \frac{M_E}{R_E^2}\). Now, solve for \(M_P\): \(M_P = \frac{8}{2} M_E = 4 M_E\).

Calculate Escape Velocity of the Other Planet

Now we have the relationship: \(R_P = 2 R_E\) and \(M_P = 4 M_E\). Use the escape velocity formula for the other planet, substitute the relationship, and simplify: \(v_{eP} = \sqrt{\frac{2G(4M_E)}{2R_E}} = \sqrt{\frac{8GM_E}{2R_E}} = \sqrt{4 \frac{2GM_E}{R_E}}\). We know that escape velocity for Earth is \(11.2 \mathrm{kms^{-1}}\), and it can be calculated from the escape velocity formula as: \(11.2 = \sqrt{\frac{2GM_E}{R_E}}\). Now substitute this value in the escape velocity equation for the other planet: \(v_{eP} = \sqrt{4 \times (11.2)^2} = 2 \times 11.2 = 22.4 \mathrm{kms^{-1}}\).

Pick the Correct Answer

The correct answer is 22.4 kms⁻¹, which corresponds to option (B).

Key concepts

Gravitational Force

Gravitational force is the attractive force between two masses. It is dictated by Newton's law of universal gravitation, which states that this force is proportional to the product of the two masses and inversely proportional to the square of the distance between them.
This can be mathematically expressed by the equation \[F = G \frac{M_1 M_2}{r^2},\]where \(F\) is the gravitational force, \(M_1\) and \(M_2\) are the masses of the objects, \(r\) is the distance between the centers of their masses, and \(G\) is the gravitational constant.

• The force decreases with increasing distance.
• It increases as the mass of either object increases.
• This force is responsible for creating the gravity we experience.

Acceleration Due to Gravity

The acceleration due to gravity is the rate at which an object speeds up when it is falling freely towards a planet. On Earth, this acceleration is approximately \(9.8 \, \text{m/s}^2\).
This value can differ on other planets depending on two main factors:

• The mass of the planet.
• The radius of the planet.

These factors are crucial because the formula for acceleration due to gravity \(g\) is \[ g = \frac{G M}{R^2}, \]where \(M\) is the mass, \(R\) is the radius of the planet, and \(G\) is the gravitational constant.
If a planet has more mass or a smaller radius, it will generally have a stronger gravitational pull, resulting in a higher \(g\) than Earth. In the original exercise, it’s mentioned that this other planet has double Earth's acceleration due to gravity, leading to implications for escape velocity.

Mass and Radius Relationship

Understanding the relationship between a planet’s mass and radius is key to predicting various gravitational phenomena. In the original exercise, you learned that doubling the diameter of a planet essentially affects its radius. Assume the planet becomes twice as wide; its radius also doubles.
Here's why this matters:

• While the radius increases, if the mass increases similarly, gravitational interactions are affected proportionally.
• If mass increases extensively, but radius increases less so, gravitational effects are amplified.

To explore this further, you can simplify the equation for gravitational force into the planet's properties and see: \[\frac{M_P}{(2R)^2} = 2\frac{M_E}{R^2}.\]Solving this helps us know how changes in size (radius) require changes in mass to maintain or alter gravitational effects. The resulting mass relationship \(M_P = 4 M_E\) shows that the other planet is significantly denser than Earth.

Planetary Motion

Planetary motion refers to the movement of planets around a star in a solar system. This motion is influenced by gravitational forces between the sun and the planet, which dictates the planet’s orbit and speed.
Escape velocity, a concept intricately related to planetary motion, is the minimum speed at which an object needs to travel to break free from a planet's gravitational pull without other propulsion. It's crucial when discussing sending rockets to space as it defines the energy required:

• The formula for escape velocity is given by \(v_e = \sqrt{\frac{2GM}{R}}\).
• Higher mass or smaller radius planets require more speed to escape their gravity.
• In the exercise, this saved us from Earth's calculations but meant recalculating for a stronger, larger planet.

For any space endeavors, accurately calculating escape velocity is essential, allowing us to understand the energy needed to reach orbit and beyond. This ensures successful spacecraft launches and confirms why the other planet's escape velocity is 22.4 \(\text{kms}^{-1}\), demonstrating the complex relationship between mass, radius, and speed.