How Does Rocket Velocity Relate to Escape Velocity and Gravitational Potential?

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The discussion centers on a physics problem involving a rocket's velocity as it coasts upward after achieving a specific speed near Earth's surface. The key equation presented is v0 = 2sqrt(gR), which leads to the conclusion that the rocket's speed far from Earth will be v = sqrt(2gR). Participants emphasize the importance of using energy conservation principles and the relationship between gravitational constants G and g. The conversation highlights a collaborative approach to problem-solving, encouraging users to engage and share their thought processes. Overall, the thread promotes learning through active participation and discussion.
K29
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Homework Statement


A rocket accelerates to speed v0 near the Earth's surface,and then coasts upwards. Show that if
v0 = 2sqrt(gR)
where R is the Earth's radius, then very far from the Earth the rockets speed will be
v = sqrt(2gR)


Homework Equations


The only equations I have thought of that may help are:
Escape Velocity:
v = sqrt(2GM/R)
Gravitational potential energy:
U= -GMm/r



The Attempt at a Solution


I have attempted using energy conservation and solving for v, however the answer appears far from that required.
This lead me to think that perhaps linear approximation may be required but I am unsure how I would apply (1+x)n = 1+nx for x<<1 in this problem.
 
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Welcome to PF!

Hi K29! Welcome to PF! :smile:

(have a square-root: √ :wink:)
K29 said:
A rocket accelerates to speed v0 near the Earth's surface,and then coasts upwards. Show that if
v0 = 2sqrt(gR)
where R is the Earth's radius, then very far from the Earth the rockets speed will be
v = sqrt(2gR)

The only equations I have thought of that may help are:
Escape Velocity:
v = sqrt(2GM/R)
Gravitational potential energy:
U= -GMm/r

Yes, those equations should do it …

but they use G, not g …

so what is the formula connecting G and g (and R)? :smile:
 
Ah yes. Thank you. All I needed to do was manipulate the final answer that I had from enegry conservations a tiny bit and then substitute that formula in

uhm as a new user,... am I meant to reply once I've got an answer??
I like the style of answering questions. Better than just copying an answer. "Help" is the way to go. At least that way I know where I've gone wrong
 
K29 said:
uhm as a new user,... am I meant to reply once I've got an answer??
I like the style of answering questions. Better than just copying an answer. "Help" is the way to go. At least that way I know where I've gone wrong

Hi K29!

Yes, it's friendlier if people reply. :smile:

Otherwise we tend to assume that they got a complete answer to the same question on another site, and never bothered with the hints from this site. :rolleyes:

The hints, btw, are the characteristic policy of PF … Greg Bernhardt :smile: reckons that members learn better if they have to do most of the work themselves!

See you around! :wink:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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