Achieving an 895 km Orbit: Calculating v

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Homework Help Overview

The discussion revolves around calculating the velocity required for a satellite to achieve an orbit at an altitude of 895 km above Earth's surface. The problem involves concepts from gravitational potential energy and kinetic energy in the context of orbital mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy at launch and potential energy in orbit, with some seeking clarification on how to substitute values into the relevant equations. Questions arise regarding the appropriate mass to use in calculations and the application of gravitational parameters.

Discussion Status

Participants are actively engaging with the problem, exploring different aspects of energy conservation in orbital mechanics. Some have provided equations and parameters, while others are seeking further clarification on the substitution process and the implications of mass in the equations.

Contextual Notes

There is a mention of a specific velocity (7404.5 m/s) that was previously calculated, which may serve as a reference point in the discussion. The use of the standard gravitational parameter for Earth is also noted, indicating a focus on precise values in the calculations.

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How fast must a satellite leave Earth's surface to reach an orbit with an altitude of 895 km?

Ek + Eg (earth) = 1/2 Eg

(could you please show how to substitute to solve for v)

this question is on the forum and it was suggested to add this v once calculated to 7404.5 m/s which was calculated from v= square root of GM/r.
 
Last edited:
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Welcome to PF.

PE + KE = PE + KE

What they want to know is the KE at launch to end in the PE and KE in orbit.
 


could you please show how to substitute for this equation? for 1/2mv squared i do not know what to use for mass.
 
Last edited:


Your PE at any point is given by -GMm/r. We will shorten that to -μ*m/r

μ = GMe is the standard gravitational parameter for Earth = 398,000 km³/s²
http://en.wikipedia.org/wiki/Standard_gravitational_parameter

This means that

-μm/Rearth + 1/2*m*V2launch = -μm/Rorbit + 1/2*m*Vorbit2

The mass of the object drops out.

V2launch = 2*(μ/Rearth -μ/Rorbit) + Vorbit2
 


Thank-you so much!
 

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