Help, I can't get out!

When an object attains escape velocity, it is unbound to the system, and has zero total energy. Therefore, the gravitational potential energy is equal in magnitude to the kinetic energy. \frac {1}{2} m v^2 + (- \frac {GMm}{r}) = 0. We can cancel the mass of the object, which we assume is small enough such that the local gravitational fields are constant along it. Shifting terms around, r= /frac {2GM}{v^2}. Simple substitution will do to find r once we put the velocity v equal to the speed of light c. Incidentally, it is to be noted that this radius is known as the Schwarzschild radius of an object with one solar mass and describes a black hole.

On the surface of the sphere, we have:

Kinetic Energy $K_1 = \dfrac{1}{2}mv^2_e$ and Potential Energy $U_1 = -\dfrac{GMm}{R}$ .

Infinitely far away from the sphere, we have:

Kinetic Energy $K_2 =0$ and Potential Energy $U_2 = 0$ .

By, conservation of mechanical energy, we have:

$U_1 + K_1 = U_2 + K_2$ $\leadsto \dfrac{1}{2}mv^2_e - \dfrac{GMm}{R} = 0$

$\leadsto R = \dfrac{2GM}{v^2_e} = \dfrac{2(6.67 \times 10^{-11} \ \text{Nm}^2\text{/kg}^2)(2 \times 10^{30} \ \text{kg})}{(3 \times 10^8 \ \text{m/s})^2} = \boxed{2964 \ \text{m}}$

Let M be mass of solid sphere Let m be mass of any body whose escape velocity is v

F=GMm/x^2

Work done in moving a body by a small distance dx

dW = F.dx dW = (GMm/x^2)dx

Integrating both sides LHS limit upper limit = W lower limit = 0

RHS limit upper limit = infinity lower limit = R

W = GMm/R

Let v be escape velocity

Kinetic Energy = (1/2)mv^2

1/2mv^2 = GMm/R v^2 = 2GM/R R= 2GM/v^2

on keeping values we get R = 2964.4km

• Method 1: You realize this problem is merely asking for the Schwarzschild radius for the sun. The formula is: $r_{s} = \frac{2Gm}{c^2}$ , where m is the mass of the sun and c is the speed of light. You plug in the numbers and you get $r_{s} = 2964 m$

• Method 2: The escape velocity is the velocity you need to move from your current position to $r = \infty$ with exactly 0 velocity left. You start with an initial kinetic energy and gravitational potential energy and end with 0 energy. Using $K = \frac{1}{2}mv^{2}$ and $U_{g} = -GmM/r$ :

$K_{initial} + U_{initial} = K_{final} + U_{final}$

$\frac{1}{2}mv^{2} + -GmM/r = \frac{1}{2}m0^{2} + -GmM/ \infty = 0$

where m is mass of your launched object, M is the mass of the sun, v is the escape velocity, and r is the distance from the center of the sun.

After manipulation:

$r = r_{s} = \frac{2GM}{v^2}$

You get the same answer $r = 2964 m$

Assuming that the escape velocity is equals to Ve=\sqrt { \frac { 2GM }{ r } } , where G is the newton's constant, M is the mass of the sphere, r is the radius of the sphere and Ve is the escape velocity. Substituting by the values given, we have that : 3×10\^ \left{ 8 \right} =\sqrt { \frac { 2×6.67×10\^ \left{ -11 \right} ×2×10\^ \left{ 30 \right} }{ r } } , and resolving the equation, whe have the answer is 2964

Let's take escape velocity as 'Ve' . Radius of sun as 'r' . Mass off the sun as 'M' . Newton's constant as 'G' . So, Ve = square root of (2 G M/r) which implies, Ve^2 = 2 G M/r which implies, r = 2 G M/Ve^2 Substituting the values we get r = (2 * 6.67 * 10 ^ -11 * 2 * 10^30) / ((3 * 10^8)^2) * =(4 * 6.67)/(9) * 10^3 * * =(2.964444) * 10^3 * * = 2964.4 m *

By using the escape velocity formula : $v_e = \sqrt{\frac{2GM}{r}}$ , we have $r = \frac{2GM}{v_e^2} = \frac{2 \times 6.67 \times 10^{-11} \times 2 \times 10^{30}}{(3 \times 10^8)^2} \approx 2964 \, \text{m}.$

Ve = root(2GM/r)

Ve = (3)10^8 m/s , G=(6.67)10^(-11) N*m^2/kg^2 and M=(2)10^30 Kg, so

(3)10^8 = root((2)(6.67)(10^(-11))(2)10^30/r)

(9)10^16 = (26.68)10^19/r

R = 2.964444...*10^3 m

Let $M$ be the mass of the sphere, $m$ be the mass of the person, $v_e$ be the escape velocity, $r$ be the radius of the sphere. By principle of conservation of energy, $(K+U_g)_i=(K+U_g)_f$ . $K_f=0$ because the final velocity is zero, $U_{gf}=0$ because its final distance is infinity. Thus, $\frac{1}{2}mv_e^2+\frac{-GMm}{r}=0+0$ . $r=\frac{2GM}{v_e^2}=\frac{(2)(6.67\times10^{-11})(2\times10^{30})}{(3\times10^8)^2}=2964.44m$ .

The total energy on the sphere is only potential(assuming no rotational kinetic energy).. The potential energy is given by U=GMm/R; where M=mass of sphere m=mass of body R=radius of sphere G=Newton's Constant An energy equal to this potential energy has to be supplied to escape out of its gravitational field which is the kinetic energy given by E=(mv^2)/2; Equating both we get it!!!

the formulae to find the radius is V = {sqrt (2gR) [escape velocity = square root over (2 x acceleration due to gravity x Radius)]}

i.e. V = sqrt [2 {(GM)/R)}] [since, g = (GM)/R^2]

or, R = (2GM)/V^2

where, G = [6.67×10^−11 Nm^2/kg^2]

M =[2 x 10^30 Kg]

V = [3 x 10^]

=2964.44

Setting the conservation of mechanical energy:

$\frac{1}{2}mv_{0}^2$ - $\frac{GMm}{r_{0}}$ = $\frac{1}{2}mv_{1}^2$ - $\frac{GMm}{r_{1}}$

We require that the projectil we fire with velocity equal to the light's from the sphere escapes from its gravitational attraction with velocity equal to 0, at infinite distance, so

$v_{0}$ = $c$

$v_{1}$ = $0$

$r_1$ $\to$ $+\infty$ $, \frac{GMm}{r_1} \to 0$

So:

$\frac{1}{2}mc^2$ - $\frac{GMm}{r_{0}}$ = $0$

Solving for $r_0$ ,

$r_0$ = $\frac{2GM}{c^2}$

which gives $r_0= 2964 m$

Please note that $r_0$ is the radius of a black hole with mass equal to the Sun's

The equation for escape velocity is $v_e = \sqrt{\frac{2GM}{r}}$ , solving for r we have $r = \frac{2GM}{v_e^2} = 2964.4$

$v_e = \sqrt {\frac {2GM}{r}}$ where $v_e$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of the sphere and $r$ is the radius of the sphere. Plugging in our information we get $r = 2.97 \times 10^3 m$ as our answer.

$v = \sqrt\frac{2GM}{r}$

putting values $r = 2964.444444444444$

We know escape velocity v can be calculated as - v = ( (2 G M) / R )^0.5 given G = 6.67×10^(−11), M = 2 10^30, v = 3 10^8

The escape velocity is the velocity such that sum of the kinetic energy and gravitational potential energy is zero. Thus, $\frac{m {v_e}^2}{2}-\frac{G M m}{r}=0$ . This provides $v_e = \sqrt{\frac{2 G M}{r}}$ , and thus, $r = \frac{2 G M}{{v_e}^2}$ . Plugging in $6.67 \times 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}$ for $G$ , $2 \times 10^{30} \text{ kg}$ for $M$ , and $3 \times 10^8 \text{ m}/\text{s}$ for $v_e$ , we get $r \approx 2964 \text{ m}$ .

$v_f^2=v_e^2+2ar=v_e^2-\frac{2Gm}{r}=0 \to r=\frac{2Gm}{v_e^2}=2964 meters$

We can relate the escape velocity to the mass and radius by the following equation: $v_e = \sqrt{ \frac{2GM}{r}}$

We simply plug in the gravitational constant $G = 6.67 \times 10^{-11}$ , mass $m = 2 \times 10^{30}$ , escape velocity $v_e = 3 \times 10^8$ and solve for the radius, which gives us: $r = \boxed{2964.44}$

Escape velocity V = (2.G.M/r)^0.5. radius r = (2.G.M/(V^2)) = 2964 meter