Problem 3 Escape velocity: (50 percent) Consider a cannon, shooting projectiles

Question

Problem 3 Escape velocity: (50 percent) Consider a cannon, shooting projectiles with speed vi, straight up in the air We assume that the Earth is spherical, and has radius Re. The force of gravity is given by where M is the mass of the Earth (look it up), m = 5.00 kg is the mass of the projectile, and G is Newton's constant (look it up) a) Explicitly compute the work done by gravity on the projectile, as it flies from Re with speed vi, to some Rf, with speed vf, and relate it to the kinetic energy b) We define the escape velocity to be the value that vi should have, such that 0 as Rf oo. What is the escape velocity for the Earth? For the Sun? c) The Schwarzschild radius of a gravitating body of mass M is the value Rs, such that had the launch radius been Rs, the escape velocity would have been equal to the speed of light. What is R, for the Earth? For the Sun? Gravitating bodies that are smaller than their Schwarzschild radius are called Black Holes". d) Going back to part a) and b). What if the projectile was not launched straight up, but at an angle? What would then be the escape velocity?

Explanation / Answer

M=5.976* 10^(24) kg, G=6.67*10^(-11) Nm2/kg2

(a)Work done by gravity can be calculated as change in Potential Energy of the body

inital PE= -(GMm)/Re ; Final PE=-(GMm)/Rf

change in PE= (GMm)*(1/Re - 1/Rf )= 1.992*10^(15)*(1/Re - 1/Rf )

This is the work done by gravity.

Now, according to work energy theorem,

Change in KE=Change in PE which is = Work done by gravity

(GMm)*(1/Re - 1/Rf ) = 0.5*m*vi2 - 0.5*m*vf2

1.992*10^(15)*(1/Re - 1/Rf ) =0.5*m*(vi2 - vf2 )

(b) Putting Rf=, 1/Rf=0, vf=0, to find vi=ve (escape velocity )

  1.992*10^(15)*(1/Re)= 0.5*m*(ve2) {(GM)*(1/Re ) = 0.5*ve2 }

putting Re=6378 km= 6378000 m

   1.992*10^(15)/6378000=0.5*5*(ve2)

ve= 11177.18 m/s= 11.177 m/s

For Sun (GM)*(1/Re ) = 0.5*ve2 can be written as

  (GMs)*(1/Rs ) = 0.5*ve2

  Ms=1.989*10^(30) kg , Rs=696000000 m

putting these values above we get

ve for sun = 617434.20 m/s = 6.174*10^(5) m/s

(c)  we know,

(GM)*(1/Re ) = 0.5*ve2

for Schwarzschild radius, Re=Rs and ve=c=3*10^(8) m/s

putting these values above we get

Rs (for earth )= 2*G*Me/c2= 8.85*10^(-3) metres

  Rs (for sun )= 2*G*Ms/c2= 2948.14 metres

(d) No change in the values of escape velocities as according to work energy theorem, change in KE is equal to change in PE as KE is a scalar quantity and does not depend upon the angle of projection.

So escape velocities value remain same irrespective of the direction of projection.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.