Answer:
The value is [tex]v = 0.996c[/tex]
Explanation:
From the question we are that
The number by which the clock on the spacecraft should be slower than that on the earth is [tex]N = 10.7[/tex]
Generally from the time dilation relation given by Albert Einstein we have that
[tex]\Delta t _1 = \frac{ \Delta t_2 }{ \sqrt{1 - \frac{v^2}{c^2 } } }[/tex]
Here [tex]\Delta t _1[/tex] is the time on earth
[tex]\Delta t _2[/tex] is the time on the spacecraft
v is the speed of the spacecraft
c is the speed of light with value [tex]c= 3.0*10^{8} \ m/s[/tex]
So
[tex]\frac{v^2 }{c^2} = 1 - \frac{\Delta t_2 }{\Delta t_1}[/tex]
Given that
[tex]\Delta t_1 = N \Delta t_2[/tex]
[tex]\Delta t_1 = 10.7 \Delta t_2[/tex]
So
[tex]v = c \sqrt{ 1 - \frac{ \Delta t_2 }{10.7 \Delta t_2} }[/tex]
[tex]v = 0.996c[/tex]