Solution 9:
Here we shall use 2 formula
1. Centripetal acceleration a = v^2/r
2. Angle of banking tan (theta) = v^2/rg
I need some help with these final problems please! 9. (a) Find the centripetal acceleration on...
During a solar eclipse the Moon is positioned directly between Earth and the Sun. Find the magnitude of the net gravitational force acting on the Moon then, due to both Earth and the Sun. The masses of the Sun, Earth, and the Moon are 1.99 x100 kg, 5.98 x1024 kg, and 7.36 x 1022 kg, respectively. The Moon's mean distance from Earth is 3.84 × 108 m, and Earth's mean distance from the Sun is 1.50 x 1011 m. The...
Mass of the Sun is 1.99 × 1030 kg, mass of the Earth is 5.98 × 1024 kg. Orbit radius of the Earth is 1.50 × 1011 m. 1) Determine the magnitude of the force of gravity between the Sun and Earth. (Express your answer to three significant figures.)
Mass of the Sun is 1.99 × × 1030 kg, mass of the Earth is 5.98 × × 1024 kg, mass of the Moon is 7.35 × × 1022 kg. Orbit radius of the Earth is 1.50 × × 1011 m, orbit radius of the Moon is 3.84 × × 108 m. 1) Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and the Sun. (Express your...
We know that the Earth moves about the Sun due to the gravity force exerted by the Sun, that is, the Earth is a “satellite” to the Sun, just like the moon to the Earth. Assume the Earth follows uniform circle motion about the Sun, and ignore gravity force due to objects other than the Sun. Estimate the orbital speed of the Earth = _______ m/s? The Earth has mass =5.98 x 1024 kg and radius 6.4 x 106 m....
Using Newton’s law of gravitation, find the centripetal acceleration of a satellite orbiting the Earth at a distance of R = 12×106 m. What is the angular velocity of that satellite? What is the period of motion? Earth’s mass: ME = 5.973×1024 kg Universal Gravitational constant: G = 6.674×10−11 m3kg−1s−2.
7) Given the mass of the sun (m,-1.99 x 1030 kg ) and the mass of the earth (me-5.98 x 1024 kg) and the distance between them (150 billion meters), find the amount of charge required on both the sun and earth to exactly balance the attractive gravitational force.
Item 13 The earth is 1.50 x 10 m from the sun. The earthD's Part A mass is 5.98 x 10 kg, while the mass of the sun is 1030 kg. 1.99 x What is earth's acceleration toward the sun? Express your answer with the approrpriate units. Units Value a= Submit Request Answer < Return to Assignment Provide Feedback
Q1. (i) (ii) (iii) Calculate the centripetal acceleration of the Earth in its orbit around the Sun, And the net force exerted on the Earth. What exerts this force on the Earth? Assume that the Earth's orbit is a circle of radius 1.5 X 101 m (iv) (v) Calculate the Gravitational Force between the Earth and the Sun Compare the Gravitational Force to that net force from section (ii) above.
(a) Explain the relationship between the universal gravitational constant G and the acceleration due to gravity at the earth's surface g. Therefore, calculate g from G using the relationships given below. Justify the choice of units for G (Nm kg?). F= mg The mass of the earth is 5.98 x 1024 kg, it's radius is Tg = 6.38 x 10 m, and G = 6.67 x 10-11 Nm?kg (10 marks) (b) Explain, including performing the integral, how the work done...
Please help me out with this assignment as soon as possible Thabk You Ch.5 -Circular Motion 6. According to Newton's Law of Gravitation, if the distance between two bodies is doubled, the attractive force between them becomes ( twice as large. Show reasoning: 0 half as large. four times as large. ( one quarter as large. 0 unchanged. fourt Kepler's third law can be simplified to T/r=1 or Tar if the orbit period T is expressed in earth years and...