# Homework Help: Energy needed to reach planets orbit and leave to leave it.

1. Jan 23, 2012

### Jacobsen

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

4,9 * 10ˇ24 kg Venus mass

T-1,9 earth days

spaceship 300 000kg

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

3. 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?

2. Jan 23, 2012

### SammyS

Staff Emeritus
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?

3. Jan 23, 2012

### Staff: Mentor

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?

4. Jan 24, 2012

### Jacobsen

T - 1,9 (164160 sec) earth days means orbiting period on the orbit. Not not necessary here.
There is also a 3)part - Lets 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?

5. Jan 24, 2012

### Staff: Mentor

Right. And I think you'll find that there should be a decimal place in that number: 1.756 kg.

6. Jan 24, 2012

### SammyS

Staff Emeritus
I believe OP uses decimal commas. 1,756 kg

7. Jan 24, 2012

### Jacobsen

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?

Last edited: Jan 24, 2012