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Your friend thinks that the escape speed should be greater for more massive objects than for less massive objects. Provide an argument for his opinion. Then provide a counterargument for why the escape speed is independent of the mass of the object. Match the words in the left column to the appropriate blanks in the sentences on the right. Terms can be used once, more than once, or not at all. Make certain each sentence is complete before submitting your answer.

User Blelump
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Answer:

Concepts and Principles

1- Kinetic Energy: The kinetic energy of an object is:

K=1/2*m*v^2 (1)

where m is the object's mass and v is its speed relative to the chosen coordinate system.

2- Gravitational potential energy of a system consisting of Earth and any object is:

U_g = -Gm_E*m_o/r*E-o (2)

where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.

Solution

The argument:

My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.

The counterargument:

We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):

K=1/2*m*v^2 (1)

and a gravitational potential energy Ug given by Equation (2):

Ug = -G*Mm/R

where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:

K+U=0

1/2*m*v^2 + (-G*Mm/R) = 0

1/2*m*v^2 = G*Mm/R

1/2*v^2 = G*M/R

solving for v we get

v = √2G*M/R

so we see v does not depend on the mass of the projectile

User MBS
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