Answer all ?'s and receive a medal :)!
v=-0.0098t+c, c is exhaust velocity in km/s, in R where c is the exhaust velocity in km/s, t is firing time, R is mass ratio. Needs to reach 7.7 km/s in order to orbit at 300 km.
1. If rocket has mass ratio of 18, exhaust velocity of 2.2 km/s and firing time is 25s. Does it make orbit?
2. What is the mass velocity in order to orbit if it has exhaust velocity of 2.5 km/s and firing time of 29s
3. Exhaust velocity- 2.4 km/s, firing time 28s, max velocity of 7.8 km/s. What is the mass ratio?

Question

Answer all ?'s and receive a medal :)! 
v=-0.0098t+c, c is exhaust velocity in km/s, in R where c is the exhaust velocity in km/s, t is firing time, R is mass ratio. Needs to reach 7.7 km/s in order to orbit at 300 km.
1. If rocket has mass ratio of 18, exhaust velocity of 2.2 km/s and firing time is 25s. Does it make orbit?
2. What is the mass velocity in order to orbit if it has exhaust velocity of 2.5 km/s and firing time of 29s 
3. Exhaust velocity- 2.4 km/s, firing time 28s, max velocity of 7.8 km/s. What is the mass ratio?

whpalmer4 · Answer

it looks to me like you didn't copy the problem exactly...

whpalmer4 · Answer

" km/s, in R where c i"  is my tipoff

whpalmer4 · Answer

"v=-0.0098t+c, c is exhaust velocity in km/s, in R where c is the exhaust velocity "

does not make sense.

anonymous · Answer

so you cant solve it?

whpalmer4 · Answer

key to solving a problem is knowing what the problem is. you've given a scrambled formula.  the correct formula is $$v = -0.0098t+c \ln R$$

whpalmer4 · Answer

problem 2 is the mass ratio, not mass velocity

anonymous · Answer

yes

whpalmer4 · Answer

Okay, I've solved all 3.  How about you? :-)

whpalmer4 · Answer

"ln" is the natural logarithm, or logarithm to the base e = 2.718281828...

anonymous · Answer

k

anonymous · Answer

did you get 6.114 for 1?

whpalmer4 · Answer

6.11382, yep.

whpalmer4 · Answer

so that's a no-go on orbit.

anonymous · Answer

im at .1225=2.5lnR what now?

whpalmer4 · Answer

how did you get that?  that's going to give you R = 1.05 or so, which is not reasonable

anonymous · Answer

can you show me how to do it?

whpalmer4 · Answer

2. What is the mass velocity in order to orbit if it has exhaust velocity of 2.5 km/s and firing time of 29s 

$$v = -0.0098t + c \ln R$$$$7.7 = -0.0098(29) + 2.5\ln R$$$$7.7 = -0.2842+2.5\ln R$$$$7.4158 = 2.5\ln R$$$$2.96632 = \ln R$$$$e^{2.96632} = R$$

whpalmer4 · Answer

hang on, let me check that...

whpalmer4 · Answer

nuts, I subtracted when I should have added!

$$7.7 = -0.2842+2.5\ln R$$$$7.9842=2.5\ln R$$$$3.19368=\ln R$$$$R=e^{3.19368}$$

anonymous · Answer

how about 3

whpalmer4 · Answer

do you have an answer to check?

anonymous · Answer

27.73

whpalmer4 · Answer

Hmm. 

3. Exhaust velocity- 2.4 km/s, firing time 28s, max velocity of 7.8 km/s. What is the mass ratio?

$$7.8 = -0.0098(28)+2.4\ln R$$$$8.0744 = 2.4\ln R$$$$3.36433=\ln R$$

whpalmer4 · Answer

that's a different result than you got.

whpalmer4 · Answer

Ah, you used 7.7 instead of 7.8

anonymous · Answer

should i have used 7.8?

whpalmer4 · Answer

Well, the problem does say "max velocity of 7.8 km/s"