Answers - Question 2

Part 1

The gravitational constant is 6.67 x 10^-11 in MKS units. Take the mass of the Earth (M) as 6.0x10²⁴ kg and the radius (R) as 6.4x10⁶ meters.

a Write down the MKS units of G.

Newton's law gives the units as ... N.m²kg^-2

b By equating the centripetal acceleration of a satellite in circular orbit with the local acceleration due to gravity find an expression for orbital velocity in terms of G and the mass of the Earth.

v²/r = GM/r²

Since the radius of the orbit is R the radius of the Earth ...

v = [GM/R]^0.5

c Show that the velocity of a ground level orbit around the earth is given by ... (gR )^0.5 where g is the acceleration due to gravity at ground level.

Since at the surface of the Earth ..

mg = GM/R²

GM = gR²

Substituting for GM in the equation above gives ...

v = [gR ]^0.5

d Hence (or otherwise) find the velocity of a ground level orbit and show that the period of such an orbit is close to 85 minutes.

v = [9.8 x 6.4x10.⁶]^0.5

= 7900 m/s

v = 2pR/T ... where T is the period.

T = [2 x 3.14 x 6.4x10⁶]/[7900 x 60]

= 85 minutes (to two significant figures)

e Explain briefly (in physical terms) why the periods of the first satellites were close to 90 minutes.
The radius of the orbit of the satellite is a little larger than the radius of the Earth and the velocity of the satellite is a little less than the velocity of a satellite in ground level orbit. Both changes give a slightly longer period.

 

Part 2

The potential energy of a mass m a distance r from a body of mass M is ...... -GMm/r.

a Where is the zero of potential energy?

The zero of potential energy is at infinity.

b By using the result obtained in Part 1 b of Question 2 (or otherwise) show that the kinetic energy of a satellite of mass m in circular orbit is half the numerical value of the potential energy.

The kinetic energy is ¹/2 mv²

= ¹/2 m[GM/r]

which is half the numerical value of ... -GMm/r

 

c Sketch a graph showing the PE, the KE, and the total energy in circular orbit as a function of orbital radius.

 

 

 

 

 

 

 

d Write down (or derive) an expression for the escape velocity of the Earth and show by numerical calculation that the escape velocity is about 11 km/s.

The escape velocity is found by making the initial KE equal to the potential energy required to reach infinity.

¹/2 mv²= GMm/R

... where v is the escape velocity and R is the radius of the Earth.

v esc = [2GM/R].^0.5

= [2x 6.67x10.^-11 x 6.0x10.²⁴(6.4x10.⁶0].^0.5

= [12.5x10.⁷].^0.5

= 11.2 km/s

 

e Find the radius (event horizon) of a hypothetical black hole with the mass of the Earth.

The event horizon is the radius from which light cannot escape. ie. the escape velocity is the velocity of light.

v esc ²= 2GM/R

R = 2GM/c.²

= [2 x 6.67x10^-11x 6.0x10²⁴]/ 9x10¹⁶

= 8.9x10^-3 meters

..................= 8.9 mm ... a remarkable result!

 

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