**Understanding Orbital Velocity and Signal Time Ratios for Mercury and the Moon***Problem Overview:*1. **Orbital Velocity of a Probe Around Mercury:** - **Given:** A probe orbits Mercury at 192 km above the surface. - **Objective:** Calculate the orbital velocity required to maintain orbit. - **Data:** - Mass of Mercury: \(3.30 \times 10^{23} \) kg - Radius of Mercury: \(2.44 \times 10^3 \) km2. **Signal Time Ratio from Earth:** - **Objective:** Find the ratio of the time it takes a signal to reach Mercury compared to the Moon. - **Distances:** - Earth to Mercury: \(57.9 \times 10^6\) km - Earth to Moon: 384,400 km3. **Wavelength Shift Calculation:** - **Given:** Signal wavelength at 15 cm. - Determine wavelength shift at the calculated orbital velocity. Assume the probe is directly moving away from Earth.**Part 1 of 4: Orbital Velocity Calculation**- **Formula:** \[ v_c = \sqrt{\frac{GM}{r}} \] - \( v_c \): Orbital velocity - \( G \): Gravitational constant - \( M \): Mass of Mercury - \( r \): Distance from the center of Mercury (radius + altitude)- **Calculation:** \[ r = 2.44 \times 10^3 \text{ km} + 192 \text{ km} = 2.63 \times 10^6 \text{ m} \] \[ v_c = \sqrt{\frac{6.67 \times 10^{-11} \times 3.30 \times 10^{23}}{2.63 \times 10^6}} = 2.89 \text{ km/s} \]**Part 2 of 4: Signal Time Ratio**- **Concept:** Signal travel time is connected to distance over speed of light. \[ t = \frac{d}{v} \]- **Expressions:** - Time to Mercury: \[ t_{\text{Mercury}} = \frac{d_{\text{Merc

Answered: Image /qna-images/answer/0f50dab2-ea72-44a0-bf22-7ccf07e036cb.jpg

You send a probe to orbit Mercury at 192 km above the surface. What orbital velocity (in km/s) is needed to keep it in orbit? (The mass of Mercury is 3.30 x 1023 kg, and the radius of Mercury is 2.44 x 103 km.) What is the ratio of the time it takes a signal from Earth to reach Mercury (d = 57.9 × 106 km) to the time it would take to reach the Moon (d = 384,400 km)? If your signal is at 15 cm, what is the wavelength shift (in cm) at this orbital velocity? (Assume the probe is at a point in its orbit in which it is moving directly away from the Earth.)

You send a probe to orbit Mercury at 192 km above the surface. What orbital velocity (in km/s) is needed to keep it in orbit? (The mass of Mercury is 3.30 x 1023 kg, and the radius of Mercury is 2.44 x 103 km.) What is the ratio of the time it takes a signal from Earth to reach Mercury (d = 57.9 × 106 km) to the time it would take to reach the Moon (d = 384,400 km)? If your signal is at 15 cm, what is the wavelength shift (in cm) at this orbital velocity? (Assume the probe is at a point in its orbit in which it is moving directly away from the Earth.)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A sphere of copper has a radius of 50.0 cm and a mass of 4690 kg. A sphere of unknown metal has a radius of 30.0 cm. The surfaces of the spheres are 20.0 cm apart. The force of gravitational attraction between the two spheres is 0.372 mN. What is the mass of the unknown metal?
Jupiter's moon, Io, loses about 1190.0 kg/s of sulfur dioxide to Jupiter's magnetosphere. At this rate, what percent of its mass would Io lose in 3.0 × 109 years? Io has a mass of about 1 × 1023 kg and there are 3.15 × 107 seconds in a a year.
It was found out that on a certain exo planet in Andromeda galaxy, the gravitational acceleration is 19.6 m/s2, and it varies as a function of elevation above the surface given by the expression g=c−bz−az2,wherec= 19.6 m/s2,b= 5.0×10−5s−2, and a= 1.0×10−2m−1s−2.Determine the height above the exo planet’s ground level where the weight of an object will decrease by 1.2 percent
A sphere of copper has a radius of 50.0 cm and a mass of 4690 kg. A sphere of unknown metal has a radius of 30.0 cm. The surfaces of the sphere are 20.0 cm apart. The force of gravitational attraction between the two spheres is 0.372 nM. What is the mass of the unknown metal?
■ ■ ay, 28 May 2023, 2:02 PM LIL 4 secs M ■ WA 45 Mag KRI Evaluate the magnitude of the net magnetic force on a current loop of THE 2 = 7.3R, 1₂ = 7.8R, and r = 2.7 R in an external magnetic field B B = 6.3 B,(-i) in terms of B. RI. Express your answer using two decimal places. Please note that a current of 41 runs on the wire. 1₂ IT FIZ Answer: 264.60 13 0 SE === 162 I I *Y ►X Two identical positively-charged particles are moving in circles within a uniform magnetic field. Particle A has a radius that is one-half that of concerning these ments
How far does light travel in one year? [This distance is known as a light-year (ly) and is used in measuring astronomical distances.] I think the answer might have to be in scientific notationn form
I-C.
How much time t does it take a radio signal to travel the 2.72 × 10¹2 m average distance from the Earth to Uranus? t = S
The mass of an electron is 9.11 x 10 ^-31 kg. The mass of a proton is 1.67 x 10 ^-27 kg. They are about 5.3×10^ −11 m apart in a hydrogen atom. What is the gravitational force between these two particles in the hydrogenatom?
The International Space Station (ISS) completes one orbit of Earth in 92 minutes. What is the radius of the orbit in kilometers (km)? You may assume the orbit is circular. The mass of the ISS is 420 kg, and the mass of the Earth is 6.0×10246.0×1024kg. Newton’s gravitational constant is 6.7×10−11N⋅m2kg26.7×10−11kg2N⋅m2.
A = 2 X 1025, B = 4.5 X 10-10, C = 3 X 10-6. AB/C = ? (2) ACB = ? (3) The Moon is approximately 400,000 km from the Earth. An atom of a certain element has a diameter of 4 X 10-8 cm. Given 1 km = 1,000 m and 1 m = 100 cm, about how many atoms of this element can be lined up between Earth and Moon? (4) A spherical planet has a radius of 2,000 km and a mass of 1025 kg. Calculate its density (mass/volume) in kilograms per cubic meter. (5) How many of the atoms in Question (3) can fit within a spherical planet with a diameter of 2 X 104 km? (6) An asteroid’s radius is 200 m and its distance from Earth is 107 km. What angle in degrees (θ) will it subtend? Use the equation θ = 57 (diameter) / distance
We will use differential equations to model the orbits and locations of Earth, Mars, and the spacecraft using Newton’s two laws mentioned above. Newton’s second law of motion in vector form is: F^→=ma^→ (1) where F^→ is the force vector in N (Newtons), and a^→ is the acceleration vector in m/s^2,and m is the mass in kg. Newton’s law of gravitation in vector form is: F^→=GMm/lr^→l*r^→/lr^→l where G=6.67x10^-11 m^3/s^2*kg is the universal gravitational constant, M is the mass of the larger object (the Sun), and is 2x10^30 kg, and m is the mass the smaller one (the planets or the spacecraft). The vector r^→ is the vector connecting the Sun to the orbiting objects. Step one ) The motion force in Equation(1), and the gravitational force in Equation(2) are equal. Equate the right hand sides of equations (1) and (2), and cancel the common factor on the left and right sides. Answer: f^→=ma^→ f=Gmm/lr^→l^2 a^→=Gmm/lr^→l^2 x r^→/lr^→l r^→=r^→/lr^→l * Gmm Could you please…