Misconceptions of 'g' ``` Name: Mary Bee Status: other Grade: other Location: Outside U.S. Country: France Date: Fall 2012 ``` Question: What G-forces, if any, did Felix Baumgartner experience when he stepped out of the capsule into the void and during his free fall, if there is a difference? Clarification Request NEWTON: We do not understand your question. Initially, the only force acting on him was the gravitational force, F = mg MARY BEE: Thank you for replying and I am sorry my question was unclear. I am not sure if I can do better but I shall try. Would it be incorrect state that: 1. a person standing motionless on the surface of the Earth experiences 1G 2. a person in free fall, outside the Earthâ€™s atmosphere with no air resistance, experiences 0G NEWTON: We are not clear on how you are using "G" Normally, G (upper case) is a universal constant (6.67 * 10^-11 N m^2/kg^2) 'g' (lower case) is the gravitational field intensity. If you are looking at a factor of the acceleration due to gravity, he was accelerating due to gravity in free fall. In #1, 2 below, you are referring to a support force (weight). In Free fall, there is no support force (this can happen within the atmosphere, too. One example might be a roller coaster.). With air resistance, there is a support force of the air resistance. In all cases, the gravitational field intensity is about the same -9.8 N/kg. MARY BEE: Thank you again and I apologise profusely for appearing to be a such a dunce. I am not sure which G it should be, upper or lower case although I am beginning to think it might be lower case because my question has something to do with the way one's weight is affected during free fall. Racing drivers experience high g-forces when accelerating and going around corners, so do jet fighter pilots when making tight turns, looping etc. sometimes even negative g when doing an inverted looping. However standing still on the ground, on Earth I always thought one would experience 1g i.e. one experiences one's own weight. Standing on the moon however one experiences only 1/6g because the moon's gravity is so much less. We weigh only 1/6 Earth weight. Now, a person jumping out of a plane is in free fall and he would therefore experience less than 1g but it cannot be 0 due to air resistance, the support force. However, Fleix Baumgartner was outside the Earth's atmosphere, so no air resistance, therefore while he was in free fall he experienced 0g until he entered the atmosphere. To put it another way, if Felix had closed his eyes standing on the ledge of the capsule, he would have felt his weight under his feet, 1g. Keeping his eyes closed he steps into the void, I would have thought he doesn't feel his weight any longer. At that moment does he not experience 0g? That's how I imagined it would be but I am beginning to doubt my reasoning. I would love to understand. Replies: 1. You are NOT a dunce. We now understand that you are caught between loose terms thrown around in the press, and precise terms used by scientists. 2. 'g' is the gravitational field intensity, usually measured in force per mass (Newtons per kilogram). This is equivalent (note: not equal) to the acceleration in the gravitational field (measured in meters per second per second). This is NOT a force. A force is a push or pull on something, acceleration is the change in velocity (how fast, fast changes). If we look at the units (I will use metric, since you are from France), force is in Newtons (kg m/s/s) and acceleration is in m/s/s These are not the same, but they are proportional at low velocities. So stating, "high g-forces when accelerating and going around corners" is confusing at best. The statement is talking about 'gravitational field intensity-forces" which just does not make sense. 3. From the context, you seem to be asking about the support force. Consider an arrow that has length and direction as a representation of force. Idealize the object in question as a dot (just to simplify the problem). Baumgartner had two equal and opposite forces acting on him just before he jumps. One is the force due to gravity down, and the other is the support force of the step acting up. This would be one arrow starting on a dot pointed upward (support force) and one arrow pointed downward (gravitational force). Since he is not accelerating, these are equal and opposite in size and direction. He feels his normal weight. When he jumps, there is only one force acting on him: the gravitational force down. This is exactly the same downward force acting on him a few moments before, when he was on the step. The gravitational field did not change, nor did his mass. Since he is no longer supported, there is a net force acting on him, causing a change in his velocity (acceleration). He no longer experiences his upward support force, so he is weightless. He still experiences his downward gravitational force. IF you are going to claim, using terms of acceleration, that at rest, he experiences 1 g, and in free fall, 0 g, that is fine, but very confusing. IF you are going to claim, using terms of gravitational field intensity, that at rest, he experiences 1g, then in free fall, he MUST experience 1 g, since the field did not change. 4. You mention measuring 'g's' for centripetal acceleration (going through curves). This is just a factor for the acceleration (not force) of the turn in increments of 9.8 m/s/s. This is just a shorthand notation to understand a situation. For example, a seated human will usually lose consciousness at an vertical acceleration of about 38 m/s/s. When engineering an amusement park ride, such as a roller coaster, it is often easier to say '4 g' than 38 m/s/s for the upward acceleration at the bottom of a drop. 5. When you speak of the moon being 1/6 g, you are on a very slippery slope. The value of the gravitational field intensity, g, varies by location in the universe. 'g' on the moon is different than 'g' on Earth. 'g' is NOT a universal constant; it changes by location. The gravitational field intensity on the surface of the moon is g, and the gravitational field intensity on Earth is g. The value at the surface of the moon is 1.6 N/kg, where the value on the surface of Earth is 9.8 N/kg. To say that the value of g on the moon is 1/6 g is not correct. That would be treating 'g' as a single value, the way pi is treated as a single value. 'g' is not a single value. I hope this helps. ---Nathan A. Unterman Click here to return to the Physics Archives

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