Gravitational Force - AP Physics C Mechanics

1. A binary star system consists of two stars of masses $M$ and $2M$ separated by distance $d$. At what distance from the star of mass $M$ is the gravitational field zero?
   Answer: $\frac{d}{1+\sqrt{2}}$ When dealing with gravitational fields from multiple sources, you're looking for the point where the net gravitational field equals zero. This occurs where the gravitational fields from both masses have equal magnitudes but opposite directions. The gravitational field from a point mass is $g = \frac{GM}{r^2}$, and it points toward the mass. For the net field to be zero, you need a point between the two stars where the fields cancel out. Let's call the distance from mass $M$ to this point $x$, so the distance from mass $2M$ is $(d-x)$. Setting the field magnitudes equal: $\frac{GM}{x^2} = \frac{G(2M)}{(d-x)^2}$ Simplifying: $\frac{1}{x^2} = \frac{2}{(d-x)^2}$ Cross-multiplying: $(d-x)^2 = 2x^2$ Taking the square root: $d-x = x\sqrt{2}$ Solving for $x$: $d = x + x\sqrt{2} = x(1+\sqrt{2})$ Therefore: $x = \frac{d}{1+\sqrt{2}}$ This confirms answer (A) is correct. Answer (B) $\frac{d}{3}$ would result from incorrectly assuming the masses are equal. Answer (C) $\frac{d}{2}$ is the midpoint, which ignores that the larger mass creates a stronger field. Answer (D) $\frac{2d}{3}$ places the zero point closer to the larger mass, which is backwards—the zero point should be closer to the smaller mass. Strategy tip: Remember that the gravitational field zero point is always closer to the smaller mass in a two-body system. Set up your equation by equating field magnitudes, not forces.
2. A thin uniform rod has mass $M$ and length $L$. A small object of mass $m$ is placed on the axis of the rod at a distance $d$ from the nearest end. Which of the following integrals correctly represents the magnitude of the gravitational force exerted by the rod on the object?
   Answer: $G \frac{Mm}{L} \int_d^{d+L} \frac{1}{x^2} dx$ To find the total force, we must integrate the forces from all infinitesimal mass elements of the rod. Let the object $m$ be at the origin. The rod lies along an axis from $x=d$ to $x=d+L$. The linear mass density is $\lambda = M/L$. A mass element is $dm = \lambda dx = (M/L)dx$. The force from this element is $dF = G \frac{m \, dm}{x^2} = G \frac{m(M/L)dx}{x^2}$. Integrating this from $x=d$ to $x=d+L$ gives the total force.
3. The magnitude of the gravitational force exerted by Earth on a satellite is $F_0$ when the satellite is at an altitude equal to Earth's radius, $R_E$. When is it appropriate to approximate the gravitational force on the satellite using $F_g = mg$, where $g$ is the acceleration due to gravity at Earth's surface?
   Answer: When the satellite is in a low Earth orbit with an altitude much less than $R_E$. The approximation $F_g = mg$, where $g$ is the surface gravity value, is valid only when the distance from Earth's center is approximately equal to Earth's radius, $R_E$. This condition is met for objects on or very near the surface. For a low Earth orbit, the altitude $h$ is much smaller than $R_E$, so $r = R_E + h \approx R_E$, and the approximation is reasonable.
4. In a laboratory on Earth, a student measures the mass of a block to be $m_g$ using a spring scale and $m_i$ using an inertial balance. The experiment is then moved to a spaceship accelerating at $a=g$ in deep space. How will the new measurements, $m'_g$ and $m'_i$, compare to the original measurements?
   Answer: $m'_g = m_g$ and $m'_i = m_i$ Mass is an intrinsic property of an object. The spring scale measures apparent weight, and in the accelerating spaceship, the apparent gravity is the same as on Earth, so it will measure the same gravitational mass. The inertial balance measures resistance to acceleration (inertial mass), which is also an intrinsic property. Therefore, both measurements will yield the same results.
5. A spherical planet has a uniform density $\rho$ and a radius $R$. Which of the following expressions represents the magnitude of the gravitational field strength $g$ at its surface?
   Answer: $\frac{4}{3} \pi G \rho R$ The gravitational field strength at the surface is $g = G \frac{M}{R^2}$. The mass of the planet is its density times its volume, $M = \rho V = \rho (\frac{4}{3} \pi R^3)$. Substituting this expression for mass into the equation for $g$ gives $g = G \frac{\rho (\frac{4}{3} \pi R^3)}{R^2} = \frac{4}{3} \pi G \rho R$.
6. A thin rod of length $L$ has a non-uniform linear mass density given by $\lambda(x) = \beta x$, where $x$ is the distance from one end at $x=0$ and $\beta$ is a positive constant. Which expression gives the magnitude of the gravitational force on a point mass $m$ located at $x=-d$?
   Answer: $G m \beta \int_0^L \frac{x}{(d+x)^2} dx$ Consider an infinitesimal segment of the rod of length $dx$ at position $x$ on the rod ($0 \le x \le L$). Its mass is $dm = \lambda(x) dx = \beta x dx$. The distance between this segment and the point mass $m$ at $x=-d$ is $r = x - (-d) = x+d$. The gravitational force exerted by this segment on $m$ is $dF = G \frac{m \, dm}{r^2} = G \frac{m (\beta x dx)}{(x+d)^2}$. To find the total force, this expression must be integrated over the entire length of the rod, from $x=0$ to $x=L$ a.
7. A uniform spherical shell has mass $M$ and radius $R$. What is the magnitude of the gravitational force exerted by the shell on a point mass $m$ located a distance $3R$ from the center of the shell?
   Answer: $G \frac{Mm}{9R^2}$ According to Newton's shell theorem, for any point outside a uniform spherical shell, the gravitational force exerted by the shell is the same as if all the shell's mass were concentrated at its center. The distance from the center is $r=3R$. Using the law of universal gravitation, $F = G \frac{Mm}{(3R)^2} = G \frac{Mm}{9R^2}$.
8. An astronaut is in a spacecraft that is accelerating away from a planet in deep space, far from any significant gravitational fields. The acceleration of the spacecraft is constant and has a magnitude of $9.8 \, \text{m/s}^2$. Which of the following best describes the astronaut's experience inside the spacecraft?
   Answer: The astronaut feels a sensation of weight similar to standing on Earth's surface. According to the principle of equivalence, the effects of a uniform gravitational field are indistinguishable from the effects of a constant, uniform acceleration. An upward acceleration of $9.8 \, \text{m/s}^2$ creates an apparent gravitational field, causing the floor of the spacecraft to push on the astronaut with a force equal to their weight on Earth. This is felt as normal weight.
9. A satellite is in a stable circular orbit around Earth. An astronaut inside the satellite releases a pen. Which statement best describes the subsequent motion of the pen?
   Answer: The pen floats at rest relative to the astronaut because both are in a state of continuous free fall around Earth. The satellite, the astronaut, and the pen are all orbiting Earth together. They are all in a state of continuous free fall, meaning they are all accelerating toward Earth at the same rate due to gravity. Because they share the same acceleration, their relative positions do not change, and the pen appears to float weightlessly next to the astronaut.
10. Planet A has radius $R$ and density $\rho$. Planet B has radius $2R$ and density $\rho/2$. Let $g_A$ and $g_B$ be the gravitational accelerations at the surfaces of Planet A and Planet B, respectively. What is the ratio $g_B / g_A$?
    Answer: $1$ The gravitational acceleration at the surface of a planet is proportional to the product of its density and radius, $g \propto \rho R$. For Planet A, $g_A \propto \rho R$. For Planet B, $g_B \propto (\rho/2)(2R) = \rho R$. Since both expressions are proportional to the same value, the surface gravities are equal, and the ratio $g_B / g_A$ is 1.
11. Two spherical objects, one with mass $M$ and the other with mass $4M$, are separated by a center-to-center distance $d$. What is the magnitude of the gravitational force exerted by the larger object on the smaller object?
    Answer: $G \frac{4M^2}{d^2}$ Newton's Law of Universal Gravitation is $F_g = G \frac{m_1 m_2}{r^2}$. Here, $m_1=M$, $m_2=4M$, and $r=d$. The force is $F_g = G \frac{(M)(4M)}{d^2} = G \frac{4M^2}{d^2}$. By Newton's third law, the force exerted by the larger object on the smaller one is equal in magnitude to the force exerted by the smaller object on the larger one.
12. An object of mass $m$ is placed inside a uniform solid sphere of mass $M$ and radius $R$ at a distance $r$ from its center, where $r < R$. The magnitude of the gravitational force on the object is proportional to which of the following quantities?
    Answer: $r$ The gravitational force inside a uniform solid sphere is given by the expression $F_g = G \frac{Mmr}{R^3}$. Since $G, M, m,$ and $R$ are constants for this situation, the force $F_g$ is directly proportional to the distance $r$ from the center.
13. A hollow spherical shell of mass $M$ and inner radius $a$ and outer radius $b$ has uniform density. A point mass $m$ is located at the geometric center of the shell. What is the gravitational force on the point mass?
    Answer: $0$ According to the shell theorem, a spherically symmetric mass distribution exerts zero net gravitational force on a point mass located at its center. This is because every element of mass in the shell has a corresponding element on the opposite side, and these paired elements exert equal and opposite forces on the central point mass, resulting in zero net force. Choices A, B, and C incorrectly apply gravitational force formulas using different characteristic distances of the shell.
14. A uniform spherical planet has mass $M$ and radius $R$. A small mass $m$ is placed at a distance $\frac{R}{2}$ below the planet's surface (at distance $\frac{R}{2}$ from the planet's center). Assuming the planet has uniform density, what is the gravitational force on mass $m$?
    Answer: $\frac{GMm}{2R^2}$ For a uniform sphere, only the mass within radius $r$ from the center contributes to the gravitational force at that point. The mass within radius $\frac{R}{2}$ is $M_{inside} = M \cdot \frac{(\frac{R}{2})^3}{R^3} = \frac{M}{8}$. The gravitational force is $F = \frac{G(\frac{M}{8})m}{(\frac{R}{2})^2} = \frac{GMm}{8} \cdot \frac{4}{R^2} = \frac{GMm}{2R^2}$. Choice A uses only the volume ratio. Choice B ignores the internal structure. Choice D incorrectly doubles the surface value.
15. Three objects, each of mass $M$, are located at the vertices of an equilateral triangle with side length $L$. What is the magnitude of the net gravitational force on one of the masses due to the other two?
    Answer: $\sqrt{3} G \frac{M^2}{L^2}$ The force on one mass is the vector sum of the forces from the other two. Each individual force has a magnitude of $F = G \frac{M^2}{L^2}$. The angle between these two force vectors is 60 degrees. Using the law of cosines for vector addition, the resultant force magnitude is $F_{net} = \sqrt{F^2 + F^2 + 2F^2 \cos(60^\circ)} = \sqrt{2F^2 + 2F^2(1/2)} = \sqrt{3F^2} = \sqrt{3} F$. Thus, the net force is $\sqrt{3} G \frac{M^2}{L^2}$.
16. A space probe travels from Earth's surface (radius $R$) to a point where the gravitational field strength is $\frac{1}{16}$ of its value at Earth's surface. At what distance from Earth's center is the probe located?
    Answer: $4R$ This question tests your understanding of how gravitational field strength varies with distance from a massive object like Earth. When you encounter gravitational field problems, always remember that the field strength follows an inverse square law with distance. The gravitational field strength at any distance $r$ from Earth's center is given by $g = \frac{GM}{r^2}$, where $G$ is the gravitational constant and $M$ is Earth's mass. At Earth's surface (distance $R$), the field strength is $g_0 = \frac{GM}{R^2}$. Since the probe experiences a field strength that is $\frac{1}{16}$ of the surface value, we can write: $\frac{GM}{r^2} = \frac{1}{16} \cdot \frac{GM}{R^2}$ Simplifying by canceling $GM$: $\frac{1}{r^2} = \frac{1}{16R^2}$ Cross-multiplying gives us $r^2 = 16R^2$, so $r = 4R$. Looking at the wrong answers: Choice B ($2R$) would give a field strength of $\frac{1}{4}$ the surface value, not $\frac{1}{16}$. Choice C ($8R$) would result in $\frac{1}{64}$ the surface strength. Choice D ($16R$) would give $\frac{1}{256}$ the surface strength. These all stem from incorrectly applying the inverse square relationship. Remember that gravitational field strength decreases as the square of the distance increases. If you want the field to be $\frac{1}{n^2}$ times weaker, you need to be $n$ times farther away. Since $\frac{1}{16} = \frac{1}{4^2}$, the distance must be $4R$.
17. A satellite orbits Earth at a distance of $2R$ from Earth's center, where $R$ is Earth's radius. If the satellite's orbital radius is increased to $4R$, by what factor does the gravitational force on the satellite change?
    Answer: Decreases by a factor of $4$ The gravitational force follows $F = \frac{GMm}{r^2}$. Initially, $F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$. Finally, $F_2 = \frac{GMm}{(4R)^2} = \frac{GMm}{16R^2}$. The ratio is $\frac{F_2}{F_1} = \frac{4R^2}{16R^2} = \frac{1}{4}$, so the force decreases by a factor of 4. Choice A uses only the linear change in radius. Choice C incorrectly cubes the radius change. Choice D incorrectly suggests an increase.
18. A person of mass $m$ stands on a scale in an elevator. The elevator is accelerating upwards with a magnitude of $g/3$. What is the reading on the scale?
    Answer: $4mg/3$ The scale reading represents the normal force $F_N$. Applying Newton's second law in the vertical direction (with up as positive): $F_{net} = F_N - mg = ma$. Given the upward acceleration $a = g/3$, the equation becomes $F_N - mg = m(g/3)$. Solving for $F_N$ gives $F_N = mg + mg/3 = 4mg/3$.
19. Which of the following experiments would be most suitable for measuring the inertial mass of an object on the International Space Station, where objects are effectively weightless?
    Answer: Attaching the object to a spring of known spring constant, displacing it, and measuring the period of oscillation. Inertial mass is a measure of an object's resistance to a change in velocity ($m = F/a$). In the effectively weightless environment of orbit, methods relying on gravitational force (weight) are not viable. An inertial balance, such as a mass-spring system, allows for the determination of inertial mass. The period of oscillation $T = 2\pi\sqrt{m/k}$ depends on the inertial mass $m$. Measuring $T$ with a known $k$ allows for the calculation of $m$.
20. A satellite of mass $m$ is in a stable circular orbit of radius $R$ around a planet of mass $M$. What is the speed of the satellite?
    Answer: $\sqrt{\frac{GM}{R}}$ For a satellite in a circular orbit, the gravitational force provides the centripetal force required to maintain the orbit. Therefore, $F_g = F_c$, which means $G\frac{Mm}{R^2} = \frac{mv^2}{R}$. The mass of the satellite $m$ cancels out. Solving for the speed $v$ yields $v^2 = \frac{GM}{R}$, so $v = \sqrt{\frac{GM}{R}}$.

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