The explosion in a cannon exerts an average force of 30000 n...

The explosion in a cannon exerts an average force of 30,000 N for L meters, the length of the cannon. What length of the cannon would be necessary to shoot a 2.0 kg projectile from the surface of the Earth to a distance of 6.84 × 10^8 m from the center of the Earth (the same as the distance to the Moon)? (G = 6.67 × 10^-11 N·m^2/kg^2, ME = 5.97 × 10^24 kg, and RE = 6.37 × 10^6 m.) Multiple Choice 3.0 km. 5.0 km. 3.9 km. 2.7 km. 4.1 km.

Transcript

00:01 Hello students in this question we are given that a cannon exerts an average force equivalent to 30 ,000 newtons for a distance equals to l meters and here the mass of the projectile is given to us which is 2 kg and we have to send it to a distance of d or here we can write this distance from the center of earth of the projectile to be 6 .84 multiplied with 10 to the power 8 meters from the center of the earth so we are also given the information that the value of capital g is equal to 6 .67 multiplied with 10 to the power minus 11 newton meter square per kg square the mass of the earth with which is represented by m .e is equals to 5 .97 multiplied with 10 to the power 24 kg and the radius of the earth is equal to 6 .37 multiplied with 10 to the power 6k .g.

01:30 For finding the length of the cannon we will firstly find the work done in sending the projectile to this much distance.

01:41 So, our formula for the gravitational potential energy is equals to minus g, m, e multiplied by the mass of the projectile divided by the distance, which is r.

01:56 So initially, the potential energy will be equals to minus 6 .67 multiplied with 10 to the hour minus 11 multiplied by the mass of the earth, which is 5 .97 multiplied with 10 to the power minus 11 multiplied by the mass of the earth, which is 5 .97 multiplied with 10 to the the power 24 multiplied by the mass of the projectile which is two divided by the initial distance so it is initially on the surface of the earth so it's distance from the center of the earth is 6 .37 multiplied with 10 to the power 6 it comes out equals to minus 6 .2512 multiplied with 10 to the power 7 joules so here it is joules.

02:56 Now u2 will be equals to minus 6 .6 7 multiplied with 10 to the power minus 11 multiplied by the mass of the earth which is 5 .97 multiplied with 10 to the power 24 and the mass of the projectile is 2 divided by the distance between them.

03:22 So the distance between them will be the radius of the earth plus the distance from the surface that is equal to 6 .84 multiplied with 10 to the power 8 so.

03:39 Here this value comes out to be minus of 5 .8 to multiply with 10 to the power 5 joules.

03:48 Now we know that the work done work done is equal to change in the potential...

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