Question

Two planets have the same mass, \(M .\) Each planet has a constant density, but the density of planet 2 is twice as high as that of planet \(1 .\) Identical objects of mass \(m\) are placed on the surfaces of the planets. What is the relationship of the gravitational potential energy on planet \(1\left(U_{1}\right)\) to that on planet \(2\left(U_{2}\right) ?\) a) \(U_{1}=U_{2}\) d) \(U_{1}=1.26 U_{2}\) b) \(U_{1}=\frac{1}{2} U_{2}\) e) \(U_{1}=0.794 U_{2}\) c) \(U_{1}=2 U_{2}\)

Short answer

(a) $U_{1} = U_{2}$ (b) $U_{1} = 2U_{2}$ (c) $U_{1} = 4U_{2}$ (d) $U_{1} = 0.5U_{2}$ (e) $U_{1} = 0.794U_{2}$ Answer: (e) $U_{1} = 0.794U_{2}$

Step-by-step solution

Recall the formula for gravitational potential energy

The gravitational potential energy of an object is given by the formula: $$U = - \frac{G m M}{r}$$ where \(G\) is the gravitational constant, \(m\) is the mass of the object, \(M\) is the mass of the planet, and \(r\) is the distance from the center of the planet to the object (essentially the radius of the planet).

Determine the relationship between the radii of the planets

Since the planets have the same mass, we can use the density formula to relate the densities and the radii. The mass of a planet is given by the formula: $$M = \rho V = \rho (4/3)\pi r^3$$ where \(\rho\) is the density and \(V\) is the volume of the planet. For planet 1, we have: $$M = \rho_1(4/3)\pi r_1^3$$ and for planet 2, the density is twice as high so we have: $$M = 2\rho_1(4/3)\pi r_2^3$$ Since both masses are equal, we can equate the two formulas and solve for the ratio of the radii: $$\rho_1(4/3)\pi r_1^3 = 2\rho_1(4/3)\pi r_2^3$$ Divide both sides by \(\rho_1(4/3)\pi\), and we get: $$r_1^3 = 2r_2^3$$ Taking the cube root of both sides gives us the relationship between the radii: $$r_1 = 2^{\frac{1}{3}}r_2$$

Calculate the ratio of the gravitational potential energies

Now we can use the formula for gravitational potential energy and the relationship between the radii to find the ratio \(U_1/U_2\): $$\frac{U_1}{U_2} = \frac{-\frac{G m M}{r_1}}{-\frac{G m M}{r_2}}$$ Simplify and substitute the expression for \(r_1\) to get: $$\frac{U_1}{U_2} = \frac{r_2}{2^{\frac{1}{3}}r_2}$$ Since the object's mass \(m\) and the planets' mass \(M\) are the same for both planets, they cancel out in the ratio, giving us: $$\frac{U_1}{U_2} = 2^{-\frac{1}{3}}$$ Plugging this into a calculator gives us an approximate value of \(0.794\).

Identify the correct answer

Comparing the calculated ratio to the available options, we see that the correct answer is: $$U_{1} = 0.794 U_{2}$$ So, the correct choice is (e).

Key concepts

Gravitational Constant

The gravitational constant, represented by the symbol \(G\), is a universal physical constant that plays a critical role in the equation of gravitational force as well as gravitational potential energy. This constant appears in Newton's law of universal gravitation and its value is approximately \(6.674 \times 10^{-11} \text{N} (\text{m/kg})^2\).

Due to its universality, the gravitational constant is the same throughout the universe, which implies that the equations of gravity can be used to understand celestial mechanics and interactions between mass bodies not only on Earth, but also across the cosmos. For example, when calculating the gravitational potential energy \(U\) of an object on or near the surface of a celestial body, like a planet, we use the formula \(U = - \frac{G m M}{r}\), where \(m\) is the mass of the object, \(M\) the mass of the celestial body, and \(r\) the separation between their centers - often incorporating the radius of the planet.

Mass-Volume Relationship

Understanding the mass-volume relationship is essential when dealing with planetary characteristics and their gravitational influence. This relationship depicts how mass (\(M\)) is distributed across the volume (\(V\)) of an object and is a direct consequence of the object's density \((\rho)\). Through the formula \(M = \rho V\), we see how the density and volume contribute to a planet's total mass.

In the context of planets or any spherical objects with a uniform density, the volume is derived from the formula for the volume of a sphere: \(V = \frac{4}{3}\pi r^3\), where \(r\) is the radius. When comparing two planets with identical masses but different densities, the planet with the higher density will have a smaller volume, and thus a smaller radius. This inverse relationship between density and radius gives us insight into how the size of a planet is not always indicative of its mass. It's necessary to apply this understanding to solve problems involving gravitational potential energy, such as the exercise in question.

Planet Density

Planet density is a fundamental attribute that determines various physical characteristics, including the strength of a planet's gravity at its surface. It's defined as the mass per unit volume of a planet and is usually expressed in \(\text{kg/m}^3\).

As planets differ in size and composition, their densities can vary significantly. For instance, a denser planet will exert stronger gravitational pull on objects at its surface than a similar mass planet with lower density. This is because the denser planet will have a smaller radius, bringing objects on the surface closer to the mass center and increasing the force of gravity according to Newton's law of gravitation.

In the exercise provided, by having the density of planet 2 being twice as much as planet 1, we infer that planet 2 is smaller in size, since it must pack the same amount of mass into a lesser volume. This in turn affects the gravitational potential energy experienced by an object on each planet's surface, resulting in the need to deeply understand the mass, volume, and density relationship when solving for the potential energy differences.