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Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1/ΔF2 is:
- a)10−2
- b)2
- c)6
- d)0.6
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Take the mean distance of the moon and the sun from the earth to be 0....
= (375)3 (4 × 10−8)
= (0.37 × 103)3(4 × 10−8)
= 1.64
Most Upvoted Answer
Take the mean distance of the moon and the sun from the earth to be 0....
G be the gravitational constant (G = 6.67430 x 10^-11 m^3 kg^-1 s^-2).
We can calculate the gravitational force between the earth and the moon using the formula:
F_moon = (G * m_earth * m_moon) / r_moon^2
where m_earth is the mass of the earth, m_moon is the mass of the moon, and r_moon is the distance between the earth and the moon.
F_moon = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (5.972 x 10^24 kg) * (8 x 10^22 kg) / (0.4 x 10^9 m)^2
F_moon = 1.193 x 10^20 N
Similarly, we can calculate the gravitational force between the earth and the sun using the same formula:
F_sun = (G * m_earth * m_sun) / r_sun^2
where m_sun is the mass of the sun and r_sun is the distance between the earth and the sun.
F_sun = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (5.972 x 10^24 kg) * (2 x 10^30 kg) / (150 x 10^9 m)^2
F_sun = 3.52 x 10^22 N
Now, let's calculate the gravitational force between the moon and the sun using the same formula:
F_moon_sun = (G * m_moon * m_sun) / (r_sun - r_moon)^2
F_moon_sun = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (8 x 10^22 kg) * (2 x 10^30 kg) / (150 x 10^9 m - 0.4 x 10^9 m)^2
F_moon_sun = 7.42 x 10^20 N
Finally, we can calculate the net gravitational force acting on the moon using the formula:
F_net = F_moon - F_moon_sun
F_net = 1.193 x 10^20 N - 7.42 x 10^20 N
F_net = -6.227 x 10^20 N
The net gravitational force acting on the moon is -6.227 x 10^20 N.
We can calculate the gravitational force between the earth and the moon using the formula:
F_moon = (G * m_earth * m_moon) / r_moon^2
where m_earth is the mass of the earth, m_moon is the mass of the moon, and r_moon is the distance between the earth and the moon.
F_moon = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (5.972 x 10^24 kg) * (8 x 10^22 kg) / (0.4 x 10^9 m)^2
F_moon = 1.193 x 10^20 N
Similarly, we can calculate the gravitational force between the earth and the sun using the same formula:
F_sun = (G * m_earth * m_sun) / r_sun^2
where m_sun is the mass of the sun and r_sun is the distance between the earth and the sun.
F_sun = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (5.972 x 10^24 kg) * (2 x 10^30 kg) / (150 x 10^9 m)^2
F_sun = 3.52 x 10^22 N
Now, let's calculate the gravitational force between the moon and the sun using the same formula:
F_moon_sun = (G * m_moon * m_sun) / (r_sun - r_moon)^2
F_moon_sun = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (8 x 10^22 kg) * (2 x 10^30 kg) / (150 x 10^9 m - 0.4 x 10^9 m)^2
F_moon_sun = 7.42 x 10^20 N
Finally, we can calculate the net gravitational force acting on the moon using the formula:
F_net = F_moon - F_moon_sun
F_net = 1.193 x 10^20 N - 7.42 x 10^20 N
F_net = -6.227 x 10^20 N
The net gravitational force acting on the moon is -6.227 x 10^20 N.
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Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest toΔF1/ΔF2 is:a)10−2b)2c)6d)0.6Correct answer is option 'B'. Can you explain this answer?
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Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest toΔF1/ΔF2 is:a)10−2b)2c)6d)0.6Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest toΔF1/ΔF2 is:a)10−2b)2c)6d)0.6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest toΔF1/ΔF2 is:a)10−2b)2c)6d)0.6Correct answer is option 'B'. Can you explain this answer?.
Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest toΔF1/ΔF2 is:a)10−2b)2c)6d)0.6Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest toΔF1/ΔF2 is:a)10−2b)2c)6d)0.6Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest toΔF1/ΔF2 is:a)10−2b)2c)6d)0.6Correct answer is option 'B'. Can you explain this answer?.
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