Gravitational Force
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AP Physics C: Mechanics › Gravitational Force
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?
$$m'_g = 0$$ and $$m'_i = 0$$
$$m'_g = m_g$$ and $$m'_i = m_i$$
$$m'_g = 0$$ and $$m'_i = m_i$$
$$m'_g = m_g$$ and $$m'_i = 0$$
Explanation
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.
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?
$$1/r^2$$
$$r^2$$
$$r$$
$$1/r$$
Explanation
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.
Planet X has mass $$M$$ and radius $$R$$. Planet Y has mass $$2M$$ and radius $$4R$$. An object of mass $$m$$ is placed on the surface of each planet. What is the ratio of the gravitational force on the object on Planet Y to the force on the object on Planet X, $$F_Y / F_X$$?
$$1/4$$
$$1/2$$
$$1/8$$
$$2$$
Explanation
The gravitational force on the surface of a planet is given by $$F = G \frac{M_{planet} m}{R_{planet}^2}$$. For Planet X, $$F_X = G \frac{M m}{R^2}$$. For Planet Y, $$F_Y = G \frac{(2M) m}{(4R)^2} = G \frac{2Mm}{16R^2} = \frac{1}{8} G \frac{Mm}{R^2}$$. The ratio is $$\frac{F_Y}{F_X} = \frac{\frac{1}{8} G \frac{Mm}{R^2}}{G \frac{Mm}{R^2}} = \frac{1}{8}$$.
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?
$$G \frac{Mm}{L} \int_d^{d+L} \frac{1}{x} dx$$
$$G \frac{Mm}{(d+L/2)^2}$$
$$G \frac{Mm}{L} \int_d^{d+L} \frac{1}{x^2} dx$$
$$G m \int_d^{d+L} \frac{1}{x^2} dx$$
Explanation
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.
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 $$x=-d$$?
$$G m \beta \int_0^L \frac{x}{(d+x)^2} dx$$
$$G m \frac{\beta L^2/2}{(d+L/2)^2}$$
$$G m \beta \int_0^L \frac{1}{(d+x)^2} dx$$
$$G m \beta \int_0^L \frac{x^2}{(d+x)^2} dx$$
Explanation
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.
A planet of uniform density has mass $$M$$ and radius $$R$$. A tunnel is drilled to a point a distance $$r = R/3$$ from the center. What is the magnitude of the gravitational force on a small object of mass $$m$$ at this location?
$$G \frac{Mm}{3R^2}$$
$$G \frac{Mm}{R^2}$$
$$G \frac{(M/27)m}{(R/3)^2}$$
$$G \frac{Mm}{9R^2}$$
Explanation
Inside a uniform solid sphere, the gravitational force at a distance $$r$$ from the center is due only to the mass enclosed within that radius, $$M_{enclosed} = M(r/R)^3$$. The force is $$F_g = G \frac{M_{enclosed} m}{r^2} = G \frac{M(r^3/R^3) m}{r^2} = G \frac{Mmr}{R^3}$$. Substituting $$r = R/3$$, the force is $$F_g = G \frac{Mm(R/3)}{R^3} = G \frac{Mm}{3R^2}$$.
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?
$$\sqrt{3} G \frac{M^2}{L^2}$$
$$G \frac{M^2}{L^2}$$
$$2 G \frac{M^2}{L^2}$$
$$\frac{\sqrt{3}}{2} G \frac{M^2}{L^2}$$
Explanation
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 $$\sqrt{3} G \frac{M^2}{L^2}$$.
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?
The pen moves in a straight line out of the satellite because there is no gravity in space.
The pen floats at rest relative to the astronaut because both are in a state of continuous free fall around Earth.
The pen is pushed to the back of the satellite because of the satellite's high orbital speed.
The pen falls to the floor of the satellite because it is closer to Earth and experiences a slightly stronger gravitational force.
Explanation
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.
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?
$$G \frac{M^2}{d^2}$$
$$G \frac{5M^2}{d^2}$$
$$G \frac{4M^2}{d^2}$$
$$G \frac{4M^2}{d}$$
Explanation
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.
A planet has a mass of $$6.0 \times 10^{24}$$ kg and a radius of $$6.4 \times 10^6$$ m. What is the magnitude of the gravitational field strength at an altitude of $$3.2 \times 10^6$$ m above the planet's surface? The universal gravitational constant is $$G = 6.67 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2$$.
$$2.4 , \text{N/kg}$$
$$39.1 , \text{N/kg}$$
$$4.3 , \text{N/kg}$$
$$9.8 , \text{N/kg}$$
Explanation
The gravitational field strength is $$g = G \frac{M}{r^2}$$. The distance $$r$$ from the center of the planet is the sum of the planet's radius and the altitude: $$r = 6.4 \times 10^6 , \text{m} + 3.2 \times 10^6 , \text{m} = 9.6 \times 10^6 , \text{m}$$. Substituting the values, $$g = (6.67 \times 10^{-11}) \frac{6.0 \times 10^{24}}{(9.6 \times 10^6)^2} \approx 4.3$$ N/kg.