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The expression Ugrav = mgh is an approximation is that is only valid near the Earth's surface. The full expression for the gravitational potential energy between two-point masses m₁, m₂ separated by distance r ES: Gm₂m₂ Here we take U = 0 at r= co. This relation also holds for spheres if we use the distance between centers. Repeat the calculation for the change in potential energy of the falling 25 kg mass with the full gravitational potential energy expression. You will want to keep six digits of precision in your calculation. Mg = 5.97 × 10²4 kg, Rg = 6,370 km. Compare your result for AU with the full expression for the gravitational potential energy with your answers for AU from the previous page using U mgh. When can you use the approximation Ugran = mgh with g = 9.8 m/s²? Choose all correct answers. If the initial point is near the Earth's surface - the final point can be anywhere. If the final point is near the Earth's surface - the initial point can be anywhere. Only if both the initial and final points are near the Earth's surface. •Pretty much never because you don't let approximations. Anytime you want since you don't care whether your satellite will reach orbit or crash.

The expression Ugrav = mgh is an approximation is that is only valid near the Earth's surface. The full expression for the gravitational potential energy between two-point masses m₁, m₂ separated by distance r ES: Gm₂m₂ Here we take U = 0 at r= co. This relation also holds for spheres if we use the distance between centers. Repeat the calculation for the change in potential energy of the falling 25 kg mass with the full gravitational potential energy expression. You will want to keep six digits of precision in your calculation. Mg = 5.97 × 10²4 kg, Rg = 6,370 km. Compare your result for AU with the full expression for the gravitational potential energy with your answers for AU from the previous page using U mgh. When can you use the approximation Ugran = mgh with g = 9.8 m/s²? Choose all correct answers. If the initial point is near the Earth's surface - the final point can be anywhere. If the final point is near the Earth's surface - the initial point can be anywhere. Only if both the initial and final points are near the Earth's surface. •Pretty much never because you don't let approximations. Anytime you want since you don't care whether your satellite will reach orbit or crash.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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d. The expression Ugrav = mgh is an approximation is that is only valid near the Earth's surface. The full expression for the gravitational potential energy between two-point masses m₁, m₂ separated by distance r is: Gm₂m₂ Ugran: = . Here we take U = 0 at r = 0o. This relation also holds for spheres if we use the distance between centers. Repeat the calculation for the change in potential energy of the falling 25 kg mass with the full gravitational potential energy expression. You will want to keep six digits of precision in your calculation. Mg = 5.97 x 1024 kg, Rg = 6,370 km. e. Compare your result for AU with the full expression for the gravitational potential energy with your answers for AU from the previous page using Umgh. f. When can you use the approximation Ugrav = mgh with g = 9.8 m/s²? Choose all correct answers. • If the initial point is near the Earth's surface - the final point can be anywhere. • If the final point is near the Earth's surface - the initial point can be anywhere. • Only if both the initial and final points are near the Earth's surface. Pretty much never because you don't let approximations. Anytime you want since you don't care whether your satellite will reach orbit or crash.
Transcribed Image Text:d. The expression Ugrav = mgh is an approximation is that is only valid near the Earth's surface. The full expression for the gravitational potential energy between two-point masses m₁, m₂ separated by distance r is: Gm₂m₂ Ugran: = . Here we take U = 0 at r = 0o. This relation also holds for spheres if we use the distance between centers. Repeat the calculation for the change in potential energy of the falling 25 kg mass with the full gravitational potential energy expression. You will want to keep six digits of precision in your calculation. Mg = 5.97 x 1024 kg, Rg = 6,370 km. e. Compare your result for AU with the full expression for the gravitational potential energy with your answers for AU from the previous page using Umgh. f. When can you use the approximation Ugrav = mgh with g = 9.8 m/s²? Choose all correct answers. • If the initial point is near the Earth's surface - the final point can be anywhere. • If the final point is near the Earth's surface - the initial point can be anywhere. • Only if both the initial and final points are near the Earth's surface. Pretty much never because you don't let approximations. Anytime you want since you don't care whether your satellite will reach orbit or crash.
1. You drop a 25 kg mass from rest in from a height of 20 km to the surface of the Earth. You want to know the kinetic energy of the object just before it hits the surface assuming that air friction is negligible (of course this is not true). You decide to use conservation of mechanical energy to determine this. a. Take h = 0 to be the surface of the Earth. Use Ugrav = mgh to determine the potential energy at the initial position and at the final position. What is the change in potential energy and what is the kinetic energy at the end? Write in numerical values in the table. Use 9.8 m/s² for g. U₁ Uf U₁ AU b. Repeat the calculation but let h = 0 be the initial position 20 km above the surface of the Earth. Determine U₁, Uf, AU, and KE, with this new choice. U₁ KEf AU KE C. Does the kinetic energy just before the object hits the Earth depend on where we choose h = 0? Explain why or why not. What quantities are important physically?
Transcribed Image Text:1. You drop a 25 kg mass from rest in from a height of 20 km to the surface of the Earth. You want to know the kinetic energy of the object just before it hits the surface assuming that air friction is negligible (of course this is not true). You decide to use conservation of mechanical energy to determine this. a. Take h = 0 to be the surface of the Earth. Use Ugrav = mgh to determine the potential energy at the initial position and at the final position. What is the change in potential energy and what is the kinetic energy at the end? Write in numerical values in the table. Use 9.8 m/s² for g. U₁ Uf U₁ AU b. Repeat the calculation but let h = 0 be the initial position 20 km above the surface of the Earth. Determine U₁, Uf, AU, and KE, with this new choice. U₁ KEf AU KE C. Does the kinetic energy just before the object hits the Earth depend on where we choose h = 0? Explain why or why not. What quantities are important physically?
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