Physics Questions:?

Question

1. a 6200 kg satellite is in circular orbit that has radius 3.3x10^7m. a net external force must act on satellite to make it change to a circular orbit that has a radius 7.0x10^m. what work must the net external force do?

2. an extreme skier starting from rest coasts down a mountain that makes an angle of 25 degrees with the horizontal, the coefficient of kinetic friction between ski and snow is 0.200. she coasts down a distance of 10.4m before coming to the edge of a cliff without slowing down, she skies off the cliff and lands downhill at a point whose vertical distance is 3.5m below the edge. how fast is she going just before she lands?

Answer 1

EDITED ANSWER Gravitational potential is G*Me*m/r. r is distance from earth center.

The work done is the difference in gravitational potential energy between the two orbits.  The potential energy of an object of mass m in an orbit of radius r is G*Me*m/r, where G = universal gravitational constant and Me is the mass of the earth.  Calculate the two values and subtract to get the energy difference, and therefore the work done.

The second question is not exactly clear.  Does she leave the cliff going downward at 25º?  or does she fly off horizontally.  In any case, her velocity at the edge of the cliff is √[2*g*L*(sinø - µ*cosø)].  This is from conservation of energy:  her potential energy at the top of the hill is m*g*sinø.  Her kinetic energy at bottom is this less the friction loss which is ¨µ*m*g*cosø * L (frictional force time distance).  The difference is the kinetic energy at the bottom, (1/2)*m*v^2.

Then resolve that v into horizontal and vertical (Vy) components.  The distance traveled vertically will be Vy*t + 0.5*g*t^2.  Solve for the time.  The vertical velocity at landing is vy + g*t, the horizontal is not changed.  The total is the sqrt of the sum of the squares.

Answer 2

1))♠ thus initial energy of sat is: Ei =E1+E2, where

E1=k*m/r1 is pot energy,

E2=0.5*m*v1^2 is kin energy of orbiting at r1=3.3e7 meter radius,

v1 is initial linear speed of sat,  

k=g*R^2 is const, g=9.8m/s^2, R=6370km is radius of Earth, m=6200kg;  

♠ thus final energy of sat is: Ef =E3+E4, where

E3=k*m/r2 is pot energy,

E4=0.5*m*v2^2 is kin energy of orbiting at r2=7.0e7m radius;

v2 is final linear speed of sat;  

♣ centrifugal force acting on sat at radius r1 is F1=m*v1^2/r1;

and centrifugal force is balance by force of gravity F1=k*m/r1^2;

thus F1=F1, thence m*v1^2/r1= k*m/r1^2, hence v1^2=k/r1;

♣ centrifugal force acting on sat at radius r2 is F2=m*v2^2/r2;

and centrifugal force is balance by gravity F2=k*m/r2^2;

thus F2=F2, thence m*v2^2/r2= k*m/r2^2, hence v2^2=k/r2;

♦ work done on the sat is Ef -Ei =(E3+E4) –(E2+E1) =

= k*m/r2 - k*m/r1 + 0.5*m*v2^2 -0.5*m*v1^2 =

= k*m*(1/r1 –1/r2) +0.5m*(k/r2 - k/r1) =

= k*m*(1/r1 –1/r2 +0.5/r1 –0.5/r2) = 1.5k*m*(1/r1 –1/r2) =

= 1.5*g*(6370e3)^2 *(1/3.3e7 –1/7.0e7) =9.554e6 J;

2))♥ Draw a pic! thus net force pushing her extremely $illy as$ forward is

F=F1-F2, where

F1=mg*sin(b) is component of her slim weight mg directed along the slope, F2=k*mg*cos(b) is force of friction, mg*cos(b) is component of her weight mg directed normal to the slope, given k=0.200, and b=25°;      

♠ net force will produce work on her pretty f***y

E=F*p, where p=10.4m is the skew distance she travels down, this will result in her kinetic energy E=0.5m*v^2, hence her speed (now directed horizontally) is u=√(2*F*p/m) = √(2*(F1-F2)*p/m) =√(2*(sinb –k*cosb)*g*p) =

= √(2*(sin25 –0.2*cos25)*9.8*10.4) = 7.01m/s;  

♣ unless the skier has broken her lovely neck so far, she’ll land with kin energy

E2=E+E1, where E1=0.5m*v^2, v is vertical component of her speed;

since her pot energy on the cliff was E1=mgh, (being h=3.5m) we get

v=√(2gh) =√(2*9.8*3.5) =8.28m/s;      

♦thus her velocity as vector is u*i –v*j;

magnitude = √(u^2 +v^2) = 10.85m/s;

not too tough! I hope you catch her under the cliff to protect her!