Help Needed: Solving Kepler's Third Law for a Solar Elliptical Orbit

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SUMMARY

The discussion focuses on solving Kepler's Third Law to determine the time required for a space probe to travel from Earth to Venus in a solar elliptical orbit. Using the provided data for Venus and Earth, the semimajor axis is calculated as 0.8615 AU. Applying Kepler's formula, the orbital period is derived as 0.800 years, which translates to approximately 292 days for the journey. The calculations confirm the probe's travel time based on the elliptical nature of the orbit.

PREREQUISITES
  • Understanding of Kepler's Laws of Planetary Motion
  • Familiarity with orbital mechanics and elliptical orbits
  • Knowledge of gravitational constants and their applications
  • Ability to perform mathematical calculations involving exponents and units of time
NEXT STEPS
  • Study Kepler's Laws in detail, focusing on their applications in celestial mechanics
  • Learn about the implications of eccentricity in elliptical orbits
  • Explore the gravitational constant (G) and its role in orbital calculations
  • Investigate the dynamics of spacecraft trajectories in solar systems
USEFUL FOR

Astronomy students, physics enthusiasts, and aerospace engineers interested in orbital mechanics and space mission planning will benefit from this discussion.

nick227
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Homework Statement



I have this problem for homework and I don't know how to even start. Can someone help? Thanks in advance.A neat and exploitive use of the sun would be to put a space probe into a solar elliptical orbit from the Earth on one side of the sun headed towards a rendezvous with Venus on the other side. Since Venus is a moving object, it would actually meet the space probe at the other end of the ellipse. So, employing Kepler's third law determine how many days are required for the space probe to travel to Venus. Consider both Venus and the Earth to have circular orbits. Also use this information:
Period(yr) Radius(AU) Eccentricity

Venus .615 .723 .007
Earth 1.000 1.000 .017

Homework Equations



t^2=((4pi^2)/(GM))a^3
t =period
G=universal gravity
M=mass of sun...?
a=semimajor axis or radius for a circle

The Attempt at a Solution



i don't really know how to start this. can anyone get me started?
 
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The farthest point from the sun in the orbit is at the radius of the Earth's orbit. The closest is at the radius of Venus' orbit on the opposite side of the sun from the first point. Draw a picture. You can use that info to find the semimajor axis of the ellipse since the distance between those two points is twice the semimajor axis.
 
so...
1.723=2a
a=.8615

t=(.8615)^(3/2)=.800yr

.800*365 = 292 days

292 is the answer. thanks a lot.
 

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