Energy needed to reach planets orbit and leave to leave it.

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Homework Help Overview

The discussion revolves around calculating the energy required for a spacecraft to ascend from the surface of Venus to a specified orbit and to escape the planet's gravitational field. The problem involves concepts from gravitational potential energy and kinetic energy in the context of celestial mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for gravitational potential energy and kinetic energy, questioning the accuracy of given answers compared to their own calculations. There is also exploration of the implications of the spacecraft's mass and the energy conversion efficiency of a nuclear energy source.

Discussion Status

Some participants express uncertainty about the correctness of the provided answers and seek clarification on the calculations. There is a recognition of potential discrepancies in the problem setup, and guidance is offered regarding the interpretation of energy values and units.

Contextual Notes

Participants note the potential confusion regarding the time period mentioned in the original post and the need for clarity on the spacecraft's potential energy at different points. There is also mention of a specific energy conversion efficiency related to nuclear energy, which adds complexity to the calculations.

Jacobsen
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How many TJ of energy is necessary to get a 300 ton spacecraft from the surface of Venus to 1) orbit at 60520km and 2) to leave the planets gravitational field?

12104km Venus diameter
6052 is Venus radius therefore.

4,9 * 10ˇ24 kg Venus mass

60520km orbit radius

T-1,9 Earth days

spaceship 300 000kg

Given answers: 14,99 TJ; 15,78 TJ

The Attempt at a Solution


1)energy at orbit -1/2Gmm/R -1/2 *(6,67*10ˇ-11)*(4,9 * 10ˇ24)*300000/60520000 (dist to orbit)= -0,81TJ
and then to get the work required i divide -0,81 TJ-(-16,2 TJ) = 15,39 TJ (the other one comes from below)
2) since binding energy=gravity ´-Gmm/R = -(6,67*10ˇ-11)*(4,9 * 10ˇ24)*300000/6052000(Venus radius) =-16,2TJ
So what exactly am calculating here? No match with given answers?
 
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Jacobsen said:
How many TJ of energy is necessary to get a 300 ton spacecraft from the surface of Venus to 1) orbit at 60520km and 2) to leave the planets gravitational field?

12104km Venus diameter
6052 is Venus radius therefore.

4,9 * 10ˇ24 kg Venus mass

60520km orbit radius

T-1,9 Earth days

spaceship 300 000kg

Given answers: 14,99 TJ; 15,78 TJ

The Attempt at a Solution


1)energy at orbit -1/2Gmm/R -1/2 *(6,67*10ˇ-11)*(4,9 * 10ˇ24)*300000/60520000 (dist to orbit)= -0,81TJ
and then to get the work required i divide -0,81 TJ-(-16,2 TJ) = 15,39 TJ (the other one comes from below)
2) since binding energy=gravity ´-Gmm/R = -(6,67*10ˇ-11)*(4,9 * 10ˇ24)*300000/6052000(Venus radius) =-16,2TJ
So what exactly am calculating here? No match with given answers?
Hello Jacobsen. Welcome to PF !

What does " T-1,9 Earth days " mean ?

What is the spacecraft 's potential energy when on the surface of Venus?

What is the spacecraft 's potential energy when on the surface of Venus?

What speed does it need to orbit with a radius of 60520 km ? What kinetic energy does this correspond to?

What is the spacecraft 's potential energy when it's in this orbit?
 
Jacobsen, it appears to me that your answers look good and that the given answers are not correct. Is there some other information that is not included in the problem description?
 
T - 1,9 (164160 sec) Earth days means orbiting period on the orbit. Not not necessary here.
There is also a 3)part - Let's assume that the starting device turns nuclear energy into mechanical at the rate of 1/10000 of fuel mass is into useful work. How much nuclear energy we need? A: 1756 kg - but i should get a different answer with my numbers right?
 
Jacobsen said:
There is also a 3)part - Let's assume that the starting device turns nuclear energy into mechanical at the rate of 1/10000 of fuel mass is into useful work. How much nuclear energy we need? A: 1756 kg - but i should get a different answer with my numbers right?

Right. And I think you'll find that there should be a decimal place in that number: 1.756 kg.
 
gneill said:
Right. And I think you'll find that there should be a decimal place in that number: 1.756 kg.
I believe OP uses decimal commas. 1,756 kg :smile:
 
1.756 kg indeed it is. My mistake, but how do i find the 1.756 kg?
It looks like i need to use the already known amount of work energy of 16,2 TJ and find a the amount of fuel per 1 J? How do i get that?
 
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