Unraveling Hubble Law and Integral: A PF Forum Guide

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    Hubble Integral Law
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SUMMARY

The discussion focuses on Hubble's Law and its application in calculating the time taken for an object to travel a specific distance in an expanding universe. The Hubble constant, denoted as H0, is approximately 1/(3.1E17 t). The user explores a simplified version of the law, where for every 10 meters, the velocity increases by 1 m/s, leading to the equation V = 0.1 D. The integral calculus approach to determine the time taken for an object 1000 meters away, moving at 100 m/s, to reach 4000 meters is also discussed, emphasizing the integration of velocity over distance.

PREREQUISITES
  • Understanding of Hubble's Law and the Hubble constant (H0)
  • Basic knowledge of integral calculus
  • Familiarity with velocity and distance formulas
  • Concept of time as a function of distance and speed
NEXT STEPS
  • Study the derivation and implications of Hubble's Law in cosmology
  • Learn advanced integral calculus techniques for physics applications
  • Explore the relationship between velocity, distance, and time in expanding universe models
  • Investigate the significance of the Hubble constant in modern astrophysics
USEFUL FOR

Astronomy students, physicists, and anyone interested in the mathematical foundations of cosmology and the dynamics of the universe.

Stephanus
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Dear PF Forum,
I know this is a very basic, basic question. But I'd like to refresh my memory.
In Hubble Law.
##V = H_0 \, D##
##H_0## is Hubble constant, aproximately ##\frac{1}{3.1E17t}##
Okayy, let's say we alter those number to an easier number.
For every 10 metres, the velocity adds 1 m/s.
This is how we write, ok?
##V = 0.1\, D##
And now this.
If an object 1000 metres from us, speeding 100m/s, of course. What time does it takes to reach 4000 metres from us?
Time is Distance / Speed.
What is the formula?
Is this correct?
##\int_a^b \frac{H_0}{x}dx, a = 1000, b = 4000##
 
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##V=dx/dt## then ##dt=dx/V## then integrate it ##V=HD## İnside the integral will be ##1/H_0xdx##
 
Last edited:
RyanH42 said:
##V=dx/dt## then ##dt=dx/V## then integrate it ##V=HD## İnside the integral will be ##1/H_0dx##
Ahh, of course H0 is the divider, how careless I am!
Thanks.
 

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