Pre-ap Review Set 4 Circular and Gravitation Answer

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A 2500kg satellite orbits 3.8*10^8m above the Earth. What is the tangential velocity of the satellite?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Correct answer:

$5.33*10^7\frac{m}{s}$

Explanation:

To solve this trouble, first recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal forcefulness acting on the satellite. That ways F_c=F_G .

Solve for F_G for the satellite. To do this, use the police force of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Think that is the distance between the centers of the 2 objects. That ways it volition be equal to the radius of the earth PLUS the orbiting distance.

Apply the given values for the masses of the objects and distance to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

At present that we know the force, we can find the acceleration. Recall that centripetal force is F_c=m*a_c . Set our 2 forces equal and solve for the centripetal acceleration.

F_G=F_c=m_s*a_c

6.87N=m_s*a_c

6.87N=2500kg*a_c

$2.7*10^{-3}\frac{m}{s^2}=a_c$

Now we tin find the tangential velocity, using the equation for centripetal dispatch. Once again, remember that the radius is equal to the sum of the radius of the Earth and the meridian of the satellite!

$a_c=\frac{v^2}{r}$

$2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}$

$(2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}$

$2.85*10^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.33*10^7\frac{m}{s}=v$

A satellite of mass orbits 3.8*10^8m above the World. If the force due to gravity of the Globe on the satellite is 7.99N , what is the mass of the satellite?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Right respond:

2914.6kg

Explanation:

To solve this problem, employ the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Recall that is the distance betwixt the centers of the 2 objects. That means it volition exist equal to the radius of the earth PLUS the orbiting distance.

Use the given values for the mass of the Earth, force of gravity, and altitude to solve for the mass of the satellite.

$7.99N=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(m_s)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$7.99N=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(m_s)(5.97*10^{24}kg)}{1.45*10^{17}m^2}$

$\frac{7.99N}{6.67*10^{-11}\frac{m^3}{kg*s^2}}=\frac{(m_s)(5.97*10^{24}kg)}{1.45*10^{17}m^2}$

$1.198*10^{11} \frac{kg^2}{m^2}=\frac{(m_s)(5.97*10^{24}kg)}{1.45*10^{17}m^2}$

$(1.198*10^{11} \frac{kg^2}{m^2})*(1.45*10^{17}m^2)={(m_s)(5.97*10^{24}kg)}$

$1.74*10^{28}kg^2={(m_s)(5.97*10^{24}kg)}$

$\frac{1.74*10^{28}kg^2}{5.97*10^{24}kg}={m_s}$

2914.6kg=m_s

A satellite of mass orbits 3.8*10^8m in a higher place the Globe. Information technology malfunctions and one-half of the satellite breaks off, leaving the satellite with merely half the original mass. What is the resulting force due to gravity on the satellite in terms of F_G , the original force due to gravity when the satellite was whole?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Correct answer:

$\frac{1}{2}F_G$

Caption:

To solve this problem, nosotros need to set up the law of universal gravitation.

When the satellite was whole:

$F_{G}_1=G\frac{m_Em_s}{r^2}$

After the satellite breaks:

$F_G_2=G\frac{m_E\frac{1}{2}m_s}{r^2}$

We tin can at present compare these two equations. Kickoff by expanding the 2d equation.

$F_G_2=G\frac{m_E\frac{1}{2}m_s}{r^2}=\frac{1}{2}*G\frac{m_Em_s}{r^2}$

We can substitute our commencement equation into the 2nd.

$F_{G}_1=G\frac{m_Em_s}{r^2}$ and $F_G_2=\frac{1}{2}*G\frac{m_Em_s}{r^2}$

$F_G_2=\frac{1}{2}*F_G_1$

This tells united states of america that the terminal force of gravity is equal to one half the original force of gravity.

A 2500kg satellite orbits 3.8*10^8m in a higher place the Earth. The satellite runs into another stationary satellite of equal mass and the two stick together. What is their resulting velocity?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Correct answer:

$2.67*10^7\frac{m}{s}$

Explanation:

Nosotros can employ the conservation of momentum to solve. Since the satellites stick together, there is only one final velocity term.

m_1v_1+m_2v_2=(m_1+m_2)v_3

We know the masses for both satellites are equal, and the second satellite is initially stationary.

$(2500kg*v_1)+(2500kg*0\frac{m}{s})=(2500kg+2500kg)v_3$

Now we need to find the velocity of the first satellite. Since the satellite is in orbit (circular motion), we need to discover the tangential velocity. Nosotros can exercise this by finding the centripetal acceleration from the centripetal force.

Recognize that the forcefulness due to gravity of the Earth on the satellite is the same as the centripetal strength interim on the satellite. That ways F_c=F_G .

Solve for F_G for the satellite. To practice this, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Recollect that is the altitude betwixt the centers of the ii objects. That means it will be equal to the radius of the earth PLUS the orbiting distance.

Utilise the given values for the masses of the objects and altitude to solve for the force of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

Now that we know the force, nosotros can find the acceleration. Remember that centripetal force is F_c=m*a_c . Set our two forces equal and solve for the centripetal acceleration.

F_G=F_c=m_s*a_c

6.87N=m_s*a_c

6.87N=2500kg*a_c

$2.7*10^{-3}\frac{m}{s^2}=a_c$

At present we can find the tangential velocity, using the equation for centripetal acceleration. Once more, remember that the radius is equal to the sum of the radius of the Globe and the peak of the satellite!

$a_c=\frac{v^2}{r}$

$2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}$

$(2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}$

$2.85*10^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.33*10^7\frac{m}{s}=v$

This value is the tangential velocity, or the initial velocity of the get-go satellite. We can plug this into the equation for conversation of momentum to solve for the final velocity of the 2 satellites.

m_1v_1+m_2v_2=(m_1+m_2)v_3

$(2500kg*5.33*10^{7}\frac{m}{s})+(2500kg*0\frac{m}{s})=(2500kg+2500kg)v_3$

$1.3325*10^{11}\frac{kg*m}{s}=(5000kg)v_3$

$\frac{1.3325*10^{11}\frac{kg*m}{s}}{5000kg}=v_3$

$2.67*10^7\frac{m}{s}=v_3$

A 2500kg satellite orbits 3.8*10^8m above the Earth. What is the menstruation of the satellite'south orbit?

$m_E=5.97*10^{24}kg$

r_E=6.37*10^5m

$G=6.67*10^{-11}\frac{m^3}{kg*s}$

Right answer:

$1.71*10^{11}s$

Explanation:

The period describes how long it takes the satellite to make one full orbit. If y'all go back to the definition of velocity, $v=\frac{\Delta x}{\Delta t}$ , we can apply that to our new circular orbit, in which the distance is equal to the circumference of the circle and the fourth dimension is equal to the menstruation: $v=\frac{2\pi r}{T}$ . The circumference divided by the period will requite u.s. the boilerplate velocity.

The problem gives us the radius, but we need to discover the tangential velocity. We tin can do this by get-go solving for the centripetal dispatch from the centripetal strength.

Recognize that the force due to gravity of the Earth on the satellite is the same as the centripetal force interim on the satellite. That means F_c=F_G .

Solve for F_G for the satellite. To practice this, use the law of universal gravitation.

$F_G=G\frac{m_1m_2}{r^2}$

Remember that is the altitude between the centers of the two objects. That ways information technology volition be equal to the radius of the earth PLUS the orbiting distance.

Utilise the given values for the masses of the objects and altitude to solve for the strength of gravity.

$F_G=G\frac{m_sm_E}{(r_E+h)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s})*\frac{(2500kg)(5.97*10^{24}kg)}{(6.37*10^5m+3.8*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.49*10^{28}kg^2}{1.45*10^{17}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*(1.03*10^{11}\frac{kg^2}{m^2})$

F_G=6.87N

Now that nosotros know the force, we can detect the acceleration. Remember that centripetal strength is F_c=m*a_c . Set our two forces equal and solve for the centripetal acceleration.

F_G=F_c=m_s*a_c

6.87N=m_s*a_c

6.87N=2500kg*a_c

$2.7*10^{-3}\frac{m}{s^2}=a_c$

Now nosotros tin can discover the tangential velocity, using the equation for centripetal acceleration. Again, remember that the radius is equal to the sum of the radius of the Earth and the height of the satellite!

$a_c=\frac{v^2}{r}$

$2.7*10^{-3}\frac{m}{s^2}=\frac{v^2}{1.45*10^{17}m}$

$(2.7*10^{-3}\frac{m}{s^2})(1.45*10^{17}m)={v^2}$

$2.85*10^{15}\frac{m^2}{s^2}=v^2$

$\sqrt{2.85*10^{15}\frac{m^2}{s^2}}=\sqrt{v^2}$

$5.33*10^7\frac{m}{s}=v$

We now have a value for the tangential velocity, which nosotros can use in the equation for velocity from the starting time to find the menstruation.

$v=\frac{\Delta x}{\Delta t}=\frac{2\pi r}{T}$

$5.33*10^7\frac{m}{s}=\frac{2\pi(1.45*10^{17}m)}{T}$

$T=\frac{2\pi(1.45*10^{17}m)}{5.33*10^7\frac{m}{s}}$

$T=1.71*10^{11}s$

2 planets are 3*10^8m apart. If the showtime planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the gravitational force between them?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Correct answer:

$1.24*10^{20}N$

Explanation:

To solve, use Newton'south law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the values for the mass of each planet, as well as the distance (radius) between them. Using these values and the gravitational abiding, we tin solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Two planets are 3*10^8m autonomously. If the first planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the acceleration on the smaller planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Correct answer:

$0.47\frac{m}{s^2}$

Explanation:

Remember that Newton'southward second law states that F=ma . The strength interim upon the planet in question will be the force due to gravity. Once we detect that, we can observe the acceleration.

To solve for the strength, employ Newton'south law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

Nosotros are given the values for the mass of each planet, as well equally the distance (radius) betwixt them. Using these values and the gravitational constant, we tin can solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

Now that we know the force of gravity, nosotros tin can utilise Newton's 2d law and the mass of the smaller planet to solve for the acceleration of gravity.

F=ma

$1.24*10^{20}N=(2.65*10^{20}kg)*a$

$\frac{1.24*10^{20}N}{2.65*10^{20}kg}=a$

$0.47\frac{m}{s^2}=a$

Ii planets are 3*10^8m autonomously. If the kickoff planet has a mass of $2.65*10^{20}kg$ and the second has a mass of $6.33*10^{26}kg$ , what is the acceleration on the larger planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Correct respond:

$1.96*10^{-7}\frac{m}{s^2}$

Explanation:

Remember that Newton's 2nd constabulary states that F=ma . The forcefulness acting upon the planet in question volition be the forcefulness due to gravity. Once we find that, nosotros tin discover the acceleration.

To solve for the force, use Newton's police force of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the values for the mass of each planet, as well as the distance (radius) betwixt them. Using these values and the gravitational abiding, we can solve for the force of gravity.

$F_G=G\frac{m_1m_2}{r^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(2.65*10^{20}kg)*(6.33*10^{26}kg)}{(3*10^8m)^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{1.67745*10^{47}kg^2}{9*10^{16}m^2}$

$F_G=(6.67*10^{-11}\frac{m^3}{kg*s^2})*1.864*10^{30}\frac{kg^2}{m^2}$

$F_G=1.24*10^{20}N$

At present that we know the force of gravity, we can use Newton's 2nd law and the mass of the larger planet to solve for the dispatch of gravity.

An astronaut has a mass of 71kg . He travels to a new planet and observes his weight is 510N on this planet'due south surface. If the radius of the planet is 6*10^7m , what is the mass of the planet?

$G=6.67*10^{-11}\frac{m^3}{kg*s^2}$

Correct answer:

$3.87*10^{26}kg$

Explanation:

To solve, use Newton's law of universal gravitation:

$F=G\frac{m_1*m_2}{r^2}$

Remember that the weight of the astronaut is the same as the gravitational strength interim between the planet and the astronaut.

$F_g=mg=G\frac{m_1*m_2}{r^2}$

We are given the gravitational constant, the radius of the planet, the mass of the astronaut, and the magnitude of the force generated. Using these values in the universal gravitation equation, we can solve for the mass of the planet.

$510N=(6.67*10^{-11}\frac{m^3}{kg*s^2})*\frac{(71kg)m_2}{(6*10^7m)^2}$

$\frac{510N}{6.67*10^{-11}\frac{m^3}{kg*s^2}}=\frac{(71kg)m_2}{(6*10^7m)^2}$

$7.65*10^{12}\frac{kg^2}{m^2}=\frac{(71kg)m_2}{3.6*10^{15}m^2}$

$(7.65*10^{12}\frac{kg^2}{m^2})(3.6*10^{15}m^2)=(71kg)m_2$

$2.75*10^{28}kg^2=(71kg)m_2$

$3.87*10^{26}=m_2$

Suppose that a person on Earth weighs 800N. If this person were to travel to a distant planet that had twice the density and the same radius of Earth, how much will the person weigh on this new planet?

$G=6.67384*10^{-11} \frac{m^{3}}{kg\cdot s^{2}}$

Right respond:

1600N

Caption:

Nosotros are given the weight of a person on Earth in units of Newtons, which means nosotros can recognize this as a force. The force that is acting on this person is the force due to gravity, which tin can exist represented past the following equation:

$F=G\frac{m_{1}m_{2}}{r^{2}}$

is the universal gravitational abiding and is equal to $6.67384*10^{-11} \frac{m^{3}}{kg\cdot s^{2}}$

$m_{1}$ is the mass of object ane

$m_{2}$ is the mass of object 2

is the distance betwixt the centers of the two objects

It's important to notation that an object'south mass will stay the aforementioned no matter where it is, but its weight will vary depending on where information technology is measured. Notice that when calculating the gravitational force, we need to consider the mass of ii objects. If nosotros set the mass of the Earth and the mass of the person in question every bit the 2 masses, nosotros tin can rewrite the equation every bit:

$F_{Earth}=G\frac{m_{person}m_{Earth}}{r^{2}}=800N$

To calculate how much the person weighs on the new planet, nosotros demand to consider the information given - that the new planet is twice every bit dense as World. This ways that for a given volume, the new planet will have twice every bit much mass every bit Earth. Furthermore, we know that the mass of the person stays the same since, every bit mentioned above, mass is constant no matter where it is measured. And, if nosotros are because a example where the volume is the same, then the distance between the centers should also be the aforementioned. Thus, nosotros can calculate the new strength as:

$F_{new\: planet}=G\frac{m_{person}(2m_{Earth})}{r^{2}}=2F_{Earth}=2\left ( 800 N\right )=1600\:N$

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Pre-ap Review Set 4 Circular and Gravitation Answer

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