What is the single best electrical ground on planet earth?

A rock is thrown on earth with an initial horizontal velocity and it hits the ground at x meters away from the?

• ground at x meters away from the point where the stone is thrown. What if: (a) The same stone is thrown with the same initial conditions but on another planet. On the new planet the thrown distance is 0.5 times the distance x on the earth. How much larger is the new planet's mass relative to earth? ( i.e new planet mass= X earth mass. Find X. ) (b) The same stone is thrown with the same initial conditions but instead on another planet with a radius 6 times the radius of the earth. With what factor will x change?

• Answer:

  (a) Assuming the radius of the new planet is the same as Earth's then the mass of the new planet is => M_np=4 M_e This is going with the assumption that the effects of the drag caused by the fluid on the new planet is just as negligible as Earth's for the purpose of this question. d_x=((v*cos(theta))/g)*((v*sin(theta))… since the stone is thrown horizontal then theta=0 therefore: d_x=((v*1)/g)*((v*0)+sqrt((v*0)^2+(2*g… since the initial velocities and initial distances above the planet do not change (and the distance above the planet is assumed to be waaaayyyy smalled than the radius of the planet) we can give those parameters a value of our choice as long as we maintain those values for each planet being compared soooo (let's not chose zero because then we won't get very far) let's set v=1 and d_y = 1 therefore d_x=(1/g)*sqrt(2*g*1)=sqrt(2)/sqrt(g) now we solve for the gravity to determine the relationship between Earth's gravity and he distance the stone travels on earth by solving for g therefore g_e = 2/d^2 now lets determine the gravity to distance traveled relationship for the new planet .5*d_x = sqrt(2)/sqrt(g) g_np=8/d^2 It's easy to see that the gravity of the new planet is 4 times the gravity of Earth The equation to determine the gavitational force experienced between two objects is F=(G*M_1*M_2)/r^2 G is the universal gravitational constant G = 6.6726 x 10^-11N-m2/kg2 and the masses of the stones thrown are equal as well as the radii of the planets therefore F_np = 4 F_e => M_np = 4 M_e (b) for this question we must either assume that this new planet 2 (np2) must be either the same mass as the first new planet or the same mass as Earth. I'm going with the 2nd new planet has a mass that is equivalent to Earth's mass F=(G*M_1*M_2)/r^2 let's simplify the problem by setting the mass of earth to 1 and the radius of earth and np2 to 1 Therefore F_e = G/1 and F_np2 = G/36 therefore F_np2 = F_e/36 d_e = sqrt(2)/sqrt(F_e) d_np2=srt(2)/sqrt(F_e/36) therefore 6 * d_np2 = d_e Could be wrong buuuttt Im pretty sure