Calculating Delta-V in Hohmann Transfer Equation

  • Thread starter Solistics
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In summary, the issue is that the value for \mu used in the calculation was incorrect, resulting in a very small delta-v of 0.00122149m/s. Once the correct value for \mu is used, the delta-v should be 4.651 x 10^{14}m^3 s^{-2}.
  • #1
Solistics
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I'm having some slight difficulties in calculating the delta-v in a hohmann transfer equation. I used the equation from wikipedia (which i hope is correct) but whenever i replace the variables i get an extremely small delta-v, is this normal?

[tex]\Delta[/tex]v=[tex]\sqrt{}(\mu/r1)[/tex]*([tex]\sqrt{}(2*r2/(r1+r2)[/tex]-1))

r1=6374000 meters
r2=405500000 meters
[tex]\mu=9.814[/tex]

my final result was 0.00122149m/s
it seems like a really small number, is this right or I'm i doing something really wrong?

i attached an excel sheet to show you my work.

thanks in advance
 

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  • #2
Solistics said:
I'm having some slight difficulties in calculating the delta-v in a hohmann transfer equation. I used the equation from wikipedia (which i hope is correct) but whenever i replace the variables i get an extremely small delta-v, is this normal?

[tex]\Delta[/tex]v=[tex]\sqrt{}(\mu/r1)[/tex]*([tex]\sqrt{}(2*r2/(r1+r2)[/tex]-1))

r1=6374000 meters
r2=405500000 meters
[tex]\mu=9.814[/tex]

my final result was 0.00122149m/s
it seems like a really small number, is this right or I'm i doing something really wrong?

i attached an excel sheet to show you my work.

thanks in advance

The problem is your value for [itex]\mu[/itex]

[tex]\mu = GM[/tex]

Where G is the universal gravitational constant and is equal to

6.67300 × 10-11 m3 kg-1 s-2

and M in this case is the mass of the Earth, or 5.97 x 1024 kg

so

4.651 X [itex]10^{14}m^3 s^{-2}[/itex]
 
  • #3
thanks for that correction.
 

Related to Calculating Delta-V in Hohmann Transfer Equation

What is the Hohmann Transfer Equation?

The Hohmann Transfer Equation is a mathematical formula used to calculate the amount of Delta-V (change in velocity) required to transfer a spacecraft from one circular orbit to another, using two engine burns at specific points along the transfer trajectory.

How is Delta-V calculated in the Hohmann Transfer Equation?

Delta-V is calculated by taking the square root of the gravitational constant (G) multiplied by the mass of the central body (M) divided by the radius of the initial orbit (r1), and then multiplied by the square root of ((2r2)/(r1+r2)), where r2 is the radius of the final orbit.

What is the purpose of the Hohmann Transfer in space travel?

The Hohmann Transfer is a fuel-efficient way to transfer a spacecraft between two circular orbits, typically used for interplanetary travel, by minimizing the amount of Delta-V required. It allows for the most efficient use of a spacecraft's limited fuel supply.

What are the assumptions made in the Hohmann Transfer Equation?

The Hohmann Transfer Equation assumes that the orbits are circular, the spacecraft is massless compared to the central body, and that all engine burns are instantaneous and occur at the appropriate points along the transfer trajectory.

What are some limitations of the Hohmann Transfer Equation?

The Hohmann Transfer Equation does not account for external forces such as atmospheric drag and solar radiation pressure, which can affect the spacecraft's trajectory. It also assumes that the orbits are coplanar, which may not always be the case in reality. Additionally, it does not take into account the time it takes to complete the transfer, which can impact the overall mission timeline.

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