MHB Speed of Perigee: Find Satellite Velocity at 750 km

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To determine the speed of a satellite at perigee, the discussion emphasizes using the conservation of energy principle, where total energy is the sum of potential and kinetic energy. The satellite's apogee speed is given as 6,800 m/s, with its apogee at 2,400 km and perigee at 750 km above Earth's surface. Participants suggest calculating potential and kinetic energy at both points to derive the satellite's speed at perigee. The gravitational parameter (μ) for Earth is noted as essential for calculations. Overall, the conversation revolves around applying these principles to find the required velocity at perigee.
cbarker1
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Dear Everybody,

How to get started for this problem?
An Earth satellite has its apogee at 2,400 km above the surface of Earth and perigee at 750 km above the surface of Earth. At apogee its speed is 6,800 m/s. What is its speed at perigee (in m/s)? Earth's radius is 6,370 km (see the figure below, which is not drawn to scale)
11-3-p-056.png

Thanks,
Cbarker1
 
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Can we apply conservation of energy?
$$E=E_p + E_k = -\frac{GMm}{r} + \frac 12 mv^2$$
 
yes. We can use the conservation of energy.
 
Cbarker1 said:
yes. We can use the conservation of energy.

Good.
So what's stopping you from getting the answer?
 
:confused:

We need to find the potential energy and the kinetic energy at both the perigee and the apogee.
That should give us an equation from which we can find the answer. (Thinking)
 
I like Serena said:
:confused:

We need to find the potential energy and the kinetic energy at both the perigee and the apogee.
That should give us an equation from which we can find the answer. (Thinking)
Do we need to find "M"?
 
Cbarker1 said:
Do we need to find "M"?

I think we need to look it up.
More specifically $\mu=GM=3.986004418(9)\times 10^{14}\text{ m}^3\text{s}^{-2}$. See wiki.
 
conservation of momentum ...

$mv_Ar_A = mv_Pr_p$
 
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