Both of the values in the numerator on the right-hand side are constants: capital □, because it’s a universal constant, and capital □, because it’s the constant mass of a large object. We’re interested in understanding from this relationship the connection between gravitational acceleration and distance □. It’s equal to a constant, the universal gravitational constant, multiplied by the mass of the much larger object divided by the distance between the two masses’ centers of mass squared. The acceleration due to gravity that our smaller mass experiences is actually independent of that mass’s value. The mass of the smaller object appears on both sides of the equation, and therefore it cancels out. Notice something interesting about this equation. If we assume that the only force acting on our smaller mass lowercase □ is the gravitational force, then we can write that that gravitational force is equal to the mass of our smaller object multiplied by its acceleration. Let’s recall that Newton’s second law of motion tells us that the net force on an object equals that object’s mass times its net acceleration. But what about the gravitational acceleration? ![]() So that’s the gravitational force that exists between these two masses. It’s equal to the universal gravitational constant, capital □, multiplied by the product of the two masses divided by the square of the distance between their centers of mass. In this case, there will be a gravitational force □ sub g between these masses. ![]() Say, for example, we have a small mass, we’ll call it lowercase □, that is a distance □ away from the center of mass of our very massive object. In general, any mass will change the gravitational field around itself and therefore tend to accelerate other massive objects. Because this object is a sphere, we can model all of its mass as though it exists at the center of the object. Let’s imagine we have here a very massive object, like a planet or a star. (E) Acceleration is proportional to the square root of □. (D) Acceleration is proportional to one over □ cubed. (□) Acceleration is proportional to one over □ squared. (B) Acceleration is proportional to one over □. Likewise, verticality of the motion and visibility of the trajectory's apex had negligible effects on the accuracy of size and distance judgments.Which of the following relations shows how the acceleration due to gravity □ around a massive object varies with the distance away from the center of mass of that object □? (A) Acceleration is proportional to □. ![]() The results showed that observers were much more sensitive to average velocity than to the gravitational acceleration pattern. Observers were asked to identify the ball that was presented and to estimate its distance. The simulated objects were balls of different diameters presented across a wide range of simulated distances. Observers viewed computer simulations of an object rising and falling on a trajectory aligned with the gravitational vector. In five experiments, we examined people's ability to utilize the size and distance information provided by gravitational acceleration. In principle, an observer could use this information to estimate the absolute size and distance of the object (Saxberg, 1987a Watson, Banks, von Hofsten, & Royden, 1992). When an object's motion is influenced by gravity, as in the rise and fall of a thrown ball, the vertical component of acceleration is roughly constant at 9.8 m/sec2.
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