Question 1

The value of g at the surface of spherical Asteroid X is 0.00600 m/s²,and the escape speed is 9.95 m/s.What is the radius of this asteroid? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The escape speed from the surface of a spherical body is given by:$$v_e = \sqrt{\frac{2GM}{r}}$$where:* $$v_e$$ is the escape speed* $$G$$ is the gravitational constant* $$M$$ is the mass of the body* $$r$$ is the radius of the bodyWe can rearrange this equation to solve for the radius:$$r = \frac{2GM}{v_e^2}$$Plugging in the given values, we get:$$r = \frac{2(6.67 	imes 10^{-11} 	ext{ N} \cdot 	ext{m}^2/	ext{kg}^2)(5.98 	imes 10^{24} 	ext{ kg})}{(9.95 	ext{ m/s})^2} = 8250 	ext{ m}$$Therefore, the radius of the asteroid is 8250 m.

Question 2

Suppose a 70-kg student were to be compressed to form a black hole.What would be the Schwarzschild radius of this black hole? (G = 6.67 × 10^-11 N ∙ m²/kg²,c = 3.0 × 10⁸ m/s)

Accepted Answer

The answer of Suppose a 70-kg student were to be...

Question 3

To what radius would the sun have to be compressed in order for it to become a black hole? The mass of the sun is 1.99 × 10³⁰ kg,G = 6.67 × 10^-11 N ∙ m²/kg²,and c = 3.00 × 10⁸ m/s

Accepted Answer

The radius to which the Sun must be compressed to form a black hole can be found using the Schwarzschild radius formula: $R = \frac{2GM}{c^2}$, where $G$ is the gravitational constant, $M$ is the mass of the Sun, and $c$ is the speed of light. Substituting the given values, $R = \frac{2(6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2)(1.99 \times 10^{30} \text{ kg})}{(3.00 \times 10^8 \text{ m/s})^2}$ yields a radius of approximately 2950 meters.

Question 4

The value of g at the surface of the spherical Asteroid X is 0.0060 m/s² and the escape velocity is 10.1 m/s.What is the mass of this asteroid? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The formula for the escape velocity of an object is:

v_esc = √(2GM/r)

where v_esc is the escape velocity, G is the gravitational constant, M is the mass of the object, and r is the radius of the object.

We can rearrange this formula to solve for the mass M:

M = (v_esc^2 * r) / (2G)

Plugging in the given values, we get:

M = (10.1^2 * 1) / (2 * 6.67 × 10^-11 * 0.0060)

M = 6.5 × 10^15 kg

Therefore, the best choice is C.

Question 5

What mass would have to be compressed to a radius of 1.0 × 10^-15 m (the order of magnitude of the radius of an atomic nucleus)in order for it to become a black hole? (G = 6.67 × 10^-11 N ∙ m²/kg²,c = 3.0 × 10⁸ m/s)

Accepted Answer

The answer of What mass would have to be compressed...

Question 6

The mass of Earth is 9.31 times the mass of Mars,and the radius of Earth is 1.88 times the radius of Mars.The escape speed for Mars is what percent of the escape speed for Earth?

Accepted Answer

The escape speed is given by:

v = sqrt((2GM)/r)

Where v is escape speed, G is gravitational constant, M is mass of the planet, and r is the radius of the planet.

Let's first find the ratio of the mass of Earth to Mars:

M_E / M_M = 9.31

Let's now find the ratio of the radius of Earth to Mars:

r_E / r_M = 1.88

Now let's use these ratios to write the escape speeds in terms of one another:

v_M / v_E = sqrt((2G(M_M / r_M))/(2G(M_E / r_E)))

v_M / v_E = sqrt((M_M / r_M)/(M_E / r_E))

v_M / v_E = sqrt((M_M / M_E)*(r_E / r_M))

Now we can substitute in the given ratios:

v_M / v_E = sqrt((1/9.31)*(1.88))

v_M / v_E = sqrt(0.202)

v_M / v_E = 0.449

Finally, we convert to a percentage:

v_M = v_E * 0.449 = 44.9% v_E

Therefore, the answer is C.

Question 7

In the past,Earth has been hit by many asteroids,some very large.Suppose one such asteroid approached Earth so that when it was very far away it was traveling at 6500 m/s along a line that would miss the center of Earth by 8,500,000 m.Because of the gravitational attraction of Earth,it departed from this straight line and curved towards Earth,colliding with it.What was the speed of this asteroid when it hit Earth if we ignore the effect of the sun? The mass of Earth is 5.97 × 10²⁴ kg,its radius is 6.38 × 10⁶ m,and G = 6.67 × 10^-11 N ∙ m²/kg².

Accepted Answer

Using conservation of energy, we can find the velocity of the asteroid just before it hits the Earth. At very far distances, the gravitational potential energy is zero and the kinetic energy is 
KE = (1/2)mv², where m is the mass of the asteroid and v is its velocity. As the asteroid approaches the Earth, potential energy is converted to kinetic energy. At the moment of impact, all the potential energy has been converted to kinetic energy. This means that 
(1/2)mv² = GMm/R, 
where M is the mass of the Earth, R is its radius, and G is the gravitational constant. Solving for v, we get 
v = sqrt(2GM/R) = sqrt(2*6.67×10⁻¹¹*5.97×10²⁴/6.38×10⁶) = 12,900 m/s. 
Therefore, the answer is B.

Question 8

A certain spherical asteroid has a mass of is 4.65 × 10¹⁷ kg and a radius of 39,600 m.If a 150-kg object is dropped from rest at an altitude of 12,400 m above the surface of this asteroid,with what speed will it hit the surface? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The answer of A certain spherical asteroid has a mass...

Question 9

Planet X-39 has a mass equal to 1/3 that of Earth and a radius equal to 1/3 that of Earth.If V is the escape speed for Earth,what is the escape speed for X-39?

Accepted Answer

The escape speed $v$ from a planet is given by the formula $v = \sqrt{\frac{2GM}{r}}$, where $G$ is the gravitational constant, $M$ is the mass of the planet, and $r$ is the radius of the planet. For planet X-39, $M = \frac{1}{3}M_{\text{Earth}}$ and $r = \frac{1}{3}r_{\text{Earth}}$. Substituting these into the escape speed formula gives $v_{X-39} = \sqrt{\frac{2G(\frac{1}{3}M_{\text{Earth}})}{\frac{1}{3}r_{\text{Earth}}}} = \sqrt{\frac{2GM_{\text{Earth}}}{r_{\text{Earth}}}} = v_{\text{Earth}}$. Therefore, the escape speed for X-39 is the same as that for Earth, which is $V$.

Question 10

Planet W-70 has a radius of 2.48 × 10⁷ m and an escape velocity of 23,300 m/s.What is the mass of W-70? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Accepted Answer

The mass of a planet can be calculated using the formula for escape velocity $v_e = \sqrt{\frac{2GM}{r}}$, where $v_e$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of the planet, and $r$ is the radius of the planet. Rearranging the formula to solve for $M$, we get $M = \frac{v_e^2 r}{2G}$. Substituting the given values ($v_e = 23300 \, \text{m/s}$, $r = 2.48 \times 10^{7} \, \text{m}$, $G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2$), we find $M = \frac{(23300)^2 \times 2.48 \times 10^{7}}{2 \times 6.67 \times 10^{-11}}$, which calculates to approximately $1.01 \times 10^{26} \, \text{kg}$.

Question 11

In 1994,fragments of Comet P/Shoemaker-Levy 9 collided with Jupiter.Assume that these fragments came from very very far away with no initial speed,and neglect the effect of the sun.What was their speed when they neared Jupiter's surface at a distance of 71.4 × 10⁶ m from its center? Jupiter has a mass of 1.90 × 10²⁷ kg,and G = 6.67 × 10^-11 N ∙ m²/kg².

Accepted Answer

The answer of In 1994,fragments of Comet P/Shoemaker-Levy 9 collided...

Quiz 7: Work and Energy

The value of g at the surface of spherical Asteroid X is 0.00600 m/s²,and the escape speed is 9.95 m/s.What is the radius of this asteroid? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Suppose a 70-kg student were to be compressed to form a black hole.What would be the Schwarzschild radius of this black hole? (G = 6.67 × 10^-11 N ∙ m²/kg²,c = 3.0 × 10⁸ m/s)

To what radius would the sun have to be compressed in order for it to become a black hole? The mass of the sun is 1.99 × 10³⁰ kg,G = 6.67 × 10^-11 N ∙ m²/kg²,and c = 3.00 × 10⁸ m/s

The value of g at the surface of the spherical Asteroid X is 0.0060 m/s² and the escape velocity is 10.1 m/s.What is the mass of this asteroid? (G = 6.67 × 10^-11 N ∙ m²/kg²)

What mass would have to be compressed to a radius of 1.0 × 10^-15 m (the order of magnitude of the radius of an atomic nucleus) in order for it to become a black hole? (G = 6.67 × 10^-11 N ∙ m²/kg²,c = 3.0 × 10⁸ m/s)

The mass of Earth is 9.31 times the mass of Mars,and the radius of Earth is 1.88 times the radius of Mars.The escape speed for Mars is what percent of the escape speed for Earth?

A certain spherical asteroid has a mass of is 4.65 × 10¹⁷ kg and a radius of 39,600 m.If a 150-kg object is dropped from rest at an altitude of 12,400 m above the surface of this asteroid,with what speed will it hit the surface? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Planet X-39 has a mass equal to 1/3 that of Earth and a radius equal to 1/3 that of Earth.If V is the escape speed for Earth,what is the escape speed for X-39?

Planet W-70 has a radius of 2.48 × 10⁷ m and an escape velocity of 23,300 m/s.What is the mass of W-70? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Quiz 7: Work and Energy

The value of g at the surface of spherical Asteroid X is 0.00600 m/s2,and the escape speed is 9.95 m/s.What is the radius of this asteroid? (G = 6.67 × 10-11 N ∙ m2/kg2)

Suppose a 70-kg student were to be compressed to form a black hole.What would be the Schwarzschild radius of this black hole? (G = 6.67 × 10-11 N ∙ m2/kg2,c = 3.0 × 108 m/s)

To what radius would the sun have to be compressed in order for it to become a black hole? The mass of the sun is 1.99 × 1030 kg,G = 6.67 × 10-11 N ∙ m2/kg2,and c = 3.00 × 108 m/s

The value of g at the surface of the spherical Asteroid X is 0.0060 m/s2 and the escape velocity is 10.1 m/s.What is the mass of this asteroid? (G = 6.67 × 10-11 N ∙ m2/kg2)

What mass would have to be compressed to a radius of 1.0 × 10-15 m (the order of magnitude of the radius of an atomic nucleus) in order for it to become a black hole? (G = 6.67 × 10-11 N ∙ m2/kg2,c = 3.0 × 108 m/s)

The mass of Earth is 9.31 times the mass of Mars,and the radius of Earth is 1.88 times the radius of Mars.The escape speed for Mars is what percent of the escape speed for Earth?

A certain spherical asteroid has a mass of is 4.65 × 1017 kg and a radius of 39,600 m.If a 150-kg object is dropped from rest at an altitude of 12,400 m above the surface of this asteroid,with what speed will it hit the surface? (G = 6.67 × 10-11 N ∙ m2/kg2)

Planet X-39 has a mass equal to 1/3 that of Earth and a radius equal to 1/3 that of Earth.If V is the escape speed for Earth,what is the escape speed for X-39?

Planet W-70 has a radius of 2.48 × 107 m and an escape velocity of 23,300 m/s.What is the mass of W-70? (G = 6.67 × 10-11 N ∙ m2/kg2)

The value of g at the surface of spherical Asteroid X is 0.00600 m/s²,and the escape speed is 9.95 m/s.What is the radius of this asteroid? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Suppose a 70-kg student were to be compressed to form a black hole.What would be the Schwarzschild radius of this black hole? (G = 6.67 × 10^-11 N ∙ m²/kg²,c = 3.0 × 10⁸ m/s)

To what radius would the sun have to be compressed in order for it to become a black hole? The mass of the sun is 1.99 × 10³⁰ kg,G = 6.67 × 10^-11 N ∙ m²/kg²,and c = 3.00 × 10⁸ m/s

The value of g at the surface of the spherical Asteroid X is 0.0060 m/s² and the escape velocity is 10.1 m/s.What is the mass of this asteroid? (G = 6.67 × 10^-11 N ∙ m²/kg²)

What mass would have to be compressed to a radius of 1.0 × 10^-15 m (the order of magnitude of the radius of an atomic nucleus) in order for it to become a black hole? (G = 6.67 × 10^-11 N ∙ m²/kg²,c = 3.0 × 10⁸ m/s)

A certain spherical asteroid has a mass of is 4.65 × 10¹⁷ kg and a radius of 39,600 m.If a 150-kg object is dropped from rest at an altitude of 12,400 m above the surface of this asteroid,with what speed will it hit the surface? (G = 6.67 × 10^-11 N ∙ m²/kg²)

Planet W-70 has a radius of 2.48 × 10⁷ m and an escape velocity of 23,300 m/s.What is the mass of W-70? (G = 6.67 × 10^-11 N ∙ m²/kg²)