A solar sail is a large, perfectly reflective sheet of lightweight material that allows a...

Question:

A solar sail is a large, perfectly reflective sheet of lightweight material that allows a spacecraft to accelerate by using the radiation pressure from the sun. A sail in the shape of a square of side 800 m is attached to a spacecraft; the mass of the spacecraft and sail is 400 kg. If the spacecraft is motionless, relative to the sun, when at a distance of 1.20 AU, what is its speed when it reaches a distance of 3.50 AU.

Gravitational Potential Energy:

Gravitational force is by nature an attractive force. And because of gravity being a force, this means that objects will start to move under its influence. Under this motion, we can induce work (or energy required to move) onto the object under the influence of the gravitational field. This is called the gravitational potential energy.

Answer and Explanation:


Given:

  • {eq}\displaystyle m = 400\ kg {/eq} is the mass of the solar sail and the spacecraft
  • {eq}\displaystyle r_1 = 1.2\ AU = 1.795\ \times\ 10^{11}\ m {/eq} is the initial distance of the craft from the sun
  • {eq}\displaystyle r_2 = 3.5\ AU = 5.236\ \times\ 10^{11}\ m {/eq} is the final distance of the craft from the sun


To solve this problem, we can use the conservation of energy. In this case, we can write:

{eq}\displaystyle K_1 + U_1 = K_2 + U_2 {/eq}

we can write the kinetic energy terms as:

{eq}\displaystyle K = \frac{1}{2} mv^2 {/eq}

and since the craft is initially motionless, we can set:

{eq}\displaystyle K_1 = 0 {/eq}


the potential energy terms can be expressed as:

{eq}\displaystyle U = \frac{GMm}{r} {/eq}

where {eq}\displaystyle M = 2\ \times\ 10^{30}\ kg {/eq} is the mass of the sun


So if we put all our information together, we get the equation:

{eq}\displaystyle \frac{GMm}{r_1} = \frac{1}{2} mv^2 + \frac{GMm}{r_2} {/eq}

we get rid of like terms:

{eq}\displaystyle \frac{GM}{r_1} = \frac{1}{2} v^2 + \frac{GM}{r_2} {/eq}

and we isolate the speed v:

{eq}\displaystyle 2\frac{GM}{r_1} + 2\frac{GM}{r_2} = v^2 {/eq}

or:

{eq}\displaystyle v = \sqrt{2GM\left(\frac{1}{r_1}-\frac{1}{r_2} \right)} {/eq}

and we substitute:

{eq}\displaystyle v = \sqrt{2(6.67\ \times\ 10^{-11}\ m^3/ kg\cdot s^2)(2\ \times\ 10^{30}\ kg)\left(\frac{1}{1.795\ \times\ 10^{11}\ m}-\frac{1}{5.236\ \times\ 10^{11}\ m} \right)} {/eq}

we will obtain the speed:

{eq}\displaystyle \boxed{v = 31,253.83\ m/s} {/eq}


Learn more about this topic:

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