Weight on the surface of a black hole ()

[lizzyb]

Homework Statement

A black hole is an object so heavy that neither matter nor even light can escape the influence of its gravitational field. Since no light can escape from it, it appears black. Suppose a mass approximately size of the Earth's mass 4.38 X 10^24 kg is packed into a small uniform sphere of radius r.

Use: The speed of light c = 2.99792 X 10^8 m/s. The universal gravitation constant G = 6.67259 X 10^-11 N m^2/kg^2.

There were two parts:

a) Findthe limiting radius r_0 when this mass becomes a black hole - easy.

b) Using Newtonian mechanics, how much would a mass of 4.64 micro-g weight at the suface of this super-dense sphere? Answer in units of N.

Homework Equations

W = g m = \frac{G M_e m}{r^2} m

The Attempt at a Solution

I did:

W = g M_o = \frac{G M_h M_o}{r^2} M_o = \frac{G M_h M_o^2}{r^2}

Using for r the value I found by the solution of (a) above, M_h, the given value of the mass of the black hole, and M_o the given value of the object (converted to kilograms). The answer was counted as wrong.

 

[joemok]

U probably used a wrong equation
F = GMm/r^2,
that is it, there should not be any square of mass

 

[lizzyb]

it's for calculating the weight, W = m g = m F = m GMm/r^2 - GMm^2/r^2

 

[OlderDan]

it's for calculating the weight, W = m g = m F = m GMm/r^2 - GMm^2/r^2

Your equation is dimensionally incorrect. mg is a force. How can mg = m*F? The problem gave you a rather precise value for the speed of light as well as the constant G. Do you think the correct calculation for part a might use those quantities? Have you learned about the Schwarzschild radius? I think that is what you need to know about to do the first part of this problem. The second part can then be done using the universal gravitaion law.