Question

(a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system if it starts at the Earth’s orbit? (b) Voyager 1 achieved a maximum speed of 125 000 km/h on its way to photograph Jupiter. Beyond what distance from the Sun is this speed sufficient to escape the solar system?

Short answer

(a) Minimum speed of the spacecraft$v_{solarescape}=42.1Km/sec$

(b) Required distance for escape $x=2.20\times {10}^{11}\mathrm{m}$

Step-by-step solution

Escape velocity

Escape velocity of earth is the minimum required velocity for any object to be free from the gravitational field of the earth. It can be represented by following formula.

$v_{escape\mathrm{velocity}}=\sqrt{\frac{2MG}{R}}$

Where G is gravitational constant, M is the mass of the object, R is the distance from the centre of the mass.

Given

$speedv=125000Km/h$

(a) Minimum speed of the spacecraft 

From the solar system the escape velocity staring at earth’s orbit will be,

$v_{solarescape}=\sqrt{\frac{2M_{sun}G}{R_{sun}}}$

$v_{solarescape}=\sqrt{\frac{2\times 1.99\times {10}^{20}kg\times 6.67\times {10}^{-27}N-m^2/Kg}{1.50\times {10}^9}}$

$v_{solarescape}=42.1Km/sec$

$v_{solarescape}=42.1Km/sec$

localid="1668087718980" $v_{solarescape}=42.1Km/sec$

(b) Required distance for escape

If we consider that x is representing the variable distance from the sun, then

Velocity will be,

$v=\frac{125000Km}{3600s}$

$v=34.7Km/sec$

On further putting all the values

$$\begin{gathered} v=\sqrt{\frac{2M_{sun}G}{x}} \\ x=\frac{2\times 1.99\times {10}^{30}kg\times 6.67\times {10}^{-11}N-m^2/Kg}{{\left(34.7Km/s\right)}^2} \\ x=2.20\times {10}^{11}m \end{gathered}$$