A spacecraft with a proper length of 400 m passes by an observer on the Earth. According to this observer, it takes 0.800 \u00b5s for the spacecraft to pass a fixed point. Determine the speed of the spacecraft as measured by the Earth-based observer. 0.857c<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n To determine the speed of the spacecraft as observed from Earth, we use the concept of length contraction<\/strong> and relative velocity<\/strong> from special relativity<\/strong>. However, in this case, we are given the proper length<\/strong> (length measured in the rest frame of the spacecraft) and the time interval<\/strong> it takes to pass a fixed point on Earth. From the Earth frame, the spacecraft is moving, so we can directly use the classical formula for speed:<\/p>\n\n\n\n We are observing from Earth, so the moving spacecraft appears contracted<\/strong>. However, in this problem, we don\u2019t<\/strong> need to apply length contraction directly, because we are given the proper length and time it takes to pass a fixed point (meaning only the front and back of the spacecraft pass the point in that time), so we can treat L0L_0 as the distance traveled in that time from Earth\u2019s point of view.<\/p>\n\n\n\n v=distancetime=L0tv = \\frac{\\text{distance}}{\\text{time}} = \\frac{L_0}{t} v=400\u2009m0.800\u00d710\u22126\u2009s=5.0\u00d7108\u2009m\/sv = \\frac{400 \\, \\text{m}}{0.800 \\times 10^{-6} \\, \\text{s}} = 5.0 \\times 10^8 \\, \\text{m\/s}<\/p>\n\n\n\n Speed of light, c=3.0\u00d7108\u2009m\/sc = 3.0 \\times 10^8 \\, \\text{m\/s} vc=5.0\u00d71083.0\u00d7108=1.67\\frac{v}{c} = \\frac{5.0 \\times 10^8}{3.0 \\times 10^8} = 1.67<\/p>\n\n\n\n But this result is not physically possible<\/strong>, since nothing can exceed the speed of light. This implies we misinterpreted<\/strong> the given length.<\/p>\n\n\n\n The proper length<\/strong> of 400 m is measured in the spacecraft\u2019s rest frame<\/strong>. To an observer on Earth, the spacecraft is contracted<\/strong> in length due to its relativistic speed. The time given (0.800 \u00b5s) is the time measured in the Earth frame<\/strong>, which is how long the contracted spacecraft takes to pass the point.<\/p>\n\n\n\n Let LL be the contracted length<\/strong> as seen from Earth. Then: L=vtL = v t<\/p>\n\n\n\n But the contracted length is also given by: L=L0\u03b3,where \u03b3=11\u2212(vc)2L = \\frac{L_0}{\\gamma}, \\quad \\text{where } \\gamma = \\frac{1}{\\sqrt{1 – \\left( \\frac{v}{c} \\right)^2 }}<\/p>\n\n\n\n So we have: vt=L0\u03b3\u21d2v=L0\u03b3tv t = \\frac{L_0}{\\gamma} \\Rightarrow v = \\frac{L_0}{\\gamma t}<\/p>\n\n\n\n Multiply both sides by \u03b3\\gamma: v\u03b3=L0tv \\gamma = \\frac{L_0}{t}<\/p>\n\n\n\n Now substitute \u03b3=11\u2212(v\/c)2\\gamma = \\frac{1}{\\sqrt{1 – (v\/c)^2}}: v\u22c511\u2212(v\/c)2=4000.800\u00d710\u22126=5.0\u00d7108v \\cdot \\frac{1}{\\sqrt{1 – (v\/c)^2}} = \\frac{400}{0.800 \\times 10^{-6}} = 5.0 \\times 10^8<\/p>\n\n\n\n Let\u2019s solve for vv numerically. Try:<\/p>\n\n\n\n Then: v=0.857\u00d73.0\u00d7108=2.571\u00d7108\u2009m\/sv = 0.857 \\times 3.0 \\times 10^8 = 2.571 \\times 10^8 \\, \\text{m\/s} \u03b3=11\u2212(0.857)2=11\u22120.734=10.266\u22481.937\\gamma = \\frac{1}{\\sqrt{1 – (0.857)^2}} = \\frac{1}{\\sqrt{1 – 0.734}} = \\frac{1}{\\sqrt{0.266}} \\approx 1.937<\/p>\n\n\n\n Now compute: v\u03b3=2.571\u00d7108\u00d71.937\u22484.98\u00d7108\u22485.0\u00d7108v \\gamma = 2.571 \\times 10^8 \\times 1.937 \\approx 4.98 \\times 10^8 \\approx 5.0 \\times 10^8<\/p>\n\n\n\n \u2705 This matches<\/strong> the required value.<\/p>\n\n\n\n v=0.857c\\boxed{v = 0.857c}<\/p>\n\n\n\n In this problem, we\u2019re analyzing how fast a spacecraft must be moving for it to pass a fixed point on Earth in a very short time, given that its proper length is 400 meters. The term proper length<\/strong> refers to the length measured in the frame in which the object (spacecraft) is at rest \u2014 in this case, 400 meters.<\/p>\n\n\n\n However, observers on Earth see the spacecraft moving at a high speed, which means they perceive it to be length-contracted<\/strong> due to the effects of special relativity. At relativistic speeds (close to the speed of light), lengths contract along the direction of motion when measured from a different inertial frame.<\/p>\n\n\n\n The observer on Earth sees the spacecraft pass by in just 0.800 microseconds. To find the speed, we must relate the proper length to the contracted length observed from Earth. Using the Lorentz transformation, we know that: L=L0\u03b3,andL=vtL = \\frac{L_0}{\\gamma}, \\quad \\text{and} \\quad L = vt<\/p>\n\n\n\n Combining the two gives: vt=L0\u03b3\u21d2v\u03b3=L0tvt = \\frac{L_0}{\\gamma} \\Rightarrow v\\gamma = \\frac{L_0}{t}<\/p>\n\n\n\n By solving this numerically and testing possible values, we find that when v=0.857cv = 0.857c, the equation is satisfied. This confirms the spacecraft is moving at 85.7% of the speed of light. This scenario illustrates how relativistic effects like time dilation and length contraction become significant at high velocities, which is a fundamental prediction of Einstein\u2019s theory of special relativity.<\/p>\n","protected":false},"excerpt":{"rendered":" A spacecraft with a proper length of 400 m passes by an observer on the Earth. According to this observer, it takes 0.800 \u00b5s for the spacecraft to pass a fixed point. Determine the speed of the spacecraft as measured by the Earth-based observer. 0.857c The correct answer and explanation is: To determine the speed […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-18630","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18630","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=18630"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18630\/revisions"}],"predecessor-version":[{"id":18632,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/18630\/revisions\/18632"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=18630"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=18630"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=18630"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Find the contracted length LL<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 2: Use speed formula<\/strong><\/h3>\n\n\n\n
\n\n\n\nStep 3: Express as a fraction of the speed of light<\/strong><\/h3>\n\n\n\n
\n\n\n\nImportant Clarification:<\/strong><\/h3>\n\n\n\n
Trial: v=0.857cv = 0.857c<\/strong><\/h3>\n\n\n\n
\n\n\n\n\u2705 Final Answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/strong><\/h3>\n\n\n\n