Question

The escape velocity from the earth is about \(11 \mathrm{kms}^{-1}\). The escape velocity from a planet having twice the radius and the same mean density as the earth is \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 22 (B) 11 (C) \(5.5\) (D) \(15.5\)

Short answer

The escape velocity of the new planet is 22 km/s.

Step-by-step solution

Recall the escape velocity formula

The formula for escape velocity is given by: \(v_e = \sqrt{\frac{2GM}{r}}\), where \(v_e\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(r\) is the radius of the planet.

Find the new planet's mass

The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^3\). Since the new planet has twice the radius of Earth, its volume will be 8 times larger. Since the new planet has the same mean density as Earth, its mass will also be 8 times larger. Let the mass of Earth be \(M_e\), the mass of the new planet is \(8M_e\).

Find the new planet's radius

The new planet's radius is twice the radius of Earth. Let the radius of Earth be \(r_e\), the radius of the new planet is \(2r_e\).

Calculate the escape velocity of the new planet

Using the escape velocity formula, substitute the mass and radius of the new planet to find its escape velocity: \(v_{e_{new}} = \sqrt{\frac{2G(8M_e)}{2r_e}}\) Notice that the \(2\) in the numerator and the \(2\) in the denominator will cancel out: \(v_{e_{new}} = \sqrt{\frac{8GM_e}{r_e}}\)

Recall the escape velocity of Earth and make the comparison

We know that the escape velocity from Earth is about 11 km/s, so: \(v_{e_{Earth}} = \sqrt{\frac{2GM_e}{r_e}} = 11\) Now, we can square both sides of this equation and obtain: \(\frac{2GM_e}{r_e} = 121\) Now, compare the escape velocity of the new planet with that of Earth: \(\frac{v_{e_{new}}}{v_{e_{Earth}}} = \frac{\sqrt{\frac{8GM_e}{r_e}}}{\sqrt{\frac{2GM_e}{r_e}}} = \sqrt{\frac{8GM_e}{r_e} \cdot \frac{r_e}{2GM_e}} = \sqrt{4} = 2\) This implies that the escape velocity of the new planet is twice that of Earth, or: \(v_{e_{new}} = 2 \cdot 11 = 22 \mathrm{kms}^{-1}\) So the correct answer is (A) 22.

Key concepts

Gravitational Constant

The gravitational constant, often denoted as \( G \), is a very important value in physics. It appears in Newton's law of universal gravitation, which shows how two objects attract each other due to gravity. Mathematically, it is represented by the formula:
\[ F = G \frac{m_1 m_2}{r^2} \]where:

• \( F \) is the force of attraction between two objects,
• \( m_1 \) and \( m_2 \) are their masses,
• \( r \) is the distance between the centers of two masses, and
• \( G \) is the gravitational constant.

The value of \( G \) is approximately \( 6.674 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\).
This constant helps determine the gravitational force exerted by massive objects like planets and stars.
Since \( G \) is universal, it applies to every mass in the universe, making it crucial for calculations of gravitational effects such as escape velocity.

Mean Density

The term "mean density" refers to the average density of an object, which in our context relates to how much mass is packed into the planet's volume.
Density, on the whole, is the amount of mass per unit volume, calculated as:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
If a planet has the same mean density as Earth, it means it has a similar distribution of mass over its volume.

• For a celestial body, this property is essential in influencing its gravitational potential.
• Even if the size of the planet changes, this density can tell us how the mass compares with that of Earth.

It's particularly useful in problems involving escape velocity as it directly ties into how the mass increases with any change in the planet's radius, a fundamental aspect of the textbook exercise.

Radius of a Planet

The radius of a planet is a fundamental geometric measure. It is the distance from the center of the planet to its surface.
For the spherical approximation that most planets follow, the radius helps in:

• Calculating the volume, using the formula for a sphere's volume \( V = \frac{4}{3}\pi r^3 \).
• Determining the gravitational pull near the planet's surface, impacting escape velocity.

In the context of the problem, knowing that the radius of the new planet is twice that of Earth tells us many things:

• The volume is \(2^3 = 8\) times bigger.
• This affects the planet's overall mass, as same mean density implies more mass with increased volume.

Thus, the radius not only governs the size but also heavily influences mass and gravitational fields.

Mass of a Planet

Mass is a measure of the amount of matter contained in an object, and for planets, it is a critical factor in gravitational studies.
When considering the mass of a planet:

• It is directly related to its volume and mean density.
• For a planet with twice the radius but same mean density as Earth, its mass becomes eight times more.

This is because the mass, \( M \), is connected as:
\[ M = \text{Density} \times \text{Volume} \]
Thus, increasing the radius influences both volume and mass.
Having a greater mass allows the planet to exert a stronger gravitational field, which is essential for determining escape velocity. Understanding this relationship between mass, radius, and density is central for computing escape velocity and analyzing planetary physics.