The straight-line distance between center of Earth and center of another planet
ID: 1481030 • Letter: T
Question
The straight-line distance between center of Earth and center of another planet s considered to be R. Imagine to have a particle with mass, m placed somewhere in this line between the Earth and this planet where the gravitational attraction of the Earth from one side is canceled by the gravitational pull of this planet from the other side. The mass of this planet is 1/36 that of the Earth. How far from the center of the Earth (in terms of R) is this point If the mass of this planet becomes 1/16 that of the Earth, where is the new location of this point (in terms of R) A 1-kg projectile is fired with initial velocity components, 20 m/s (horizontally) and 40 m/s (vertically upward) from a point on the earth s surface. The kinetic energy of the projectile when it reaches the highest point in its trajectory is The work done on firing the projectile isExplanation / Answer
Solution 4) Let us consider that the force of attraction from the boh the eart and the planet are equal .
Therfore, mMe / r12 = m Mp / r22
where ,m = mass of the point in between the line joining the earth and the planet.
Me = mass of the earth
Mp = mass of the planet = (1/36) *Me
r1 = distance of the point from the earth
r2 = distance of the point from the planet
R= r1 + r2
hence, (m * Me) / r12 = (m * Mp) / r22
36= r12/r22
r2 = r1 / 6
therefore, r1 + r1 / 6 =R
r1 =6R/7
This is the distance between point and the centre of mass of the earth.
b)
let Mp = (1/16)* Me
Me / r12 = (1/16)Me/r22
r2 = r1/4
therfore,r1 =4R/5
solution 5)
a) K.E at the highest point = (1/2)*m Vx2 = (1/2 ) *1*400 =200 J
b) work done = (1/2)m*(Vx2 + Vy2) = (1/2)*1*(20*20 + 40*40)=1000 J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.