Question

Suppose a rocket ship in deep space moves with constant acceleration equal to 9.8 m/s2, which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at 3.0 × 108 m/s? (b) How far will it travel in so doing?

Answer 1

errr your way off

Answer 2

lol much simpler than that

Answer 3

a is in metres and b is in seconds for the answers

Answer 4

sorry you have misinterpreted the question

Answer 5

whats the answer?

Answer 6

a) 10.6m
(b)41.5s

Answer 7

i have no idea about the working lol

Answer 8

The acceleration is constant, and =g, in the moving (accelerating) frame, which means that
dα/dτ = g/c,

where β = v/c = tanh α
α = the velocity parameter
τ = proper time
g = acceleration of gravity at Earth's surface =~9.80 m/s^2
v = velocity
c = speed of light in vacuum =~3.00*10^8 m/s
tanh is the hyperbolic tangent, defined in terms of hyperbolic sine and cosine:
tanh x = sinh x / cosh x;
where
sinh x = (e^x - e^-x)/2
cosh x = (e^x + e^-x)/2, so that

tanh x = (e^x - e^-x)/(e^x + e^-x)
= (e^2x - 1)/(e^2x + 1)
= 1 - 2/(e^2x + 1)

(a)

This actually has two answers -- elapsed time aboard ship (τ), and elapsed time back on Earth (t)

So, because at t=0, τ, v, β, and α are all =0,
dα/dτ = g/c means that
α = gτ/c

For v = 0.12c, β = 0.12, α = arctanh β = 0.12058, so
τ = cα/g = 3.691*10^6 s = 42.72 days

=======================PROPER TIME====================

=6 weeks + 17 hr, as experienced in the spaceship (cause it's proper time, which goes with the moving object)

To get elapsed time, t, in the "Earth" frame, use the definition of proper time in terms of "Earth" spacetime coordinates, t and x:
dτ^2 = dt^2 - (dx/c)^2;
dτ = √(1 - β^2)dt = dt/γ
where γ = 1/√(1 - β^2) = cosh α
dα/dτ = g/c then becomes
g/c = γdα/dt = cosh α dα/dt
the solution of which is
sinh α = gt/c; ---> from which we get:
t = (c/g)sinh α = γv/g = v/(g√(1 - β^2)) = 3.700*10^6 s = 42.83 days
===============COORDINATE TIME ON EARTH===================

=~6 weeks + 20 hr, as seen on Earth

(b)

To get distance traveled (this has to be in the "Earth" frame):
α = arcsinh(gt/c)
β = tanh α = sinh α/cosh α
= sinh α/√[1 + (sinh α)^2]
= gt/√[c^2 + (gt)^2]
v = dx/dt = cβ = cgt/√[c^2 + (gt)^2],
which we integrate, and use the fact that x=0 when t=0 (that's where the "-1" comes from at the end of the next line):

x = (c^2/g)(√[1 + (gt/c)^2] - 1)
= 6.6845*10^13

Answer 9

OHHH

Answer 10

sorry i was looking at the wrong question window...ignore my last replies...however..

Answer 11

(a) 3.1 × 106 s; (b) 4.6 × 1013 m

Answer 12

10^6 not 106

Answer 13

10^13

Answer 14

τ = cα/g = 3.691*10^6 s

Answer 15

For distance i get x= 6.6845*10^13m

Answer 16

hmm ok this is just a physics problem (motion in 1d)

Answer 17

it can be done just using constant acceleration formulas....don't need trig

Answer 18

1st year uni stuff...maybe high school stuff

Answer 19

Without the relativistic (SR) realm, using newtonian kinematics, this question can be done too,
vf = 0.12 * 3 * 10^8 m/sec
Given v0 = 0 and
a = 9.8 m/sec^2

At constant acceleration,
vf = v0 + at
vf = 9.8 m/sec^2 t
Equate the two speeds
vf = 9.8t = 0.12 ( 3 ) x10^8 m/sec.
t = 3.673469.3x10^6 secs.

for (b)

From the time computed above, compute the distance at constant acceleration and v0 = 0 using the distance formula:
s = 1/2 at^2
s = 6,612x10^13 m.

Answer 20

ah k answers still different

Answer 21

i'm getting it from a textbook :p

Answer 22

thanks for trying