1. The distance between the Sun and Mercury varies from dmin = 4.600 107 km to d
ID: 1335919 • Letter: 1
Question
1. The distance between the Sun and Mercury varies from dmin = 4.600 107 km to dmax = 6.982 107 km over the course of Mercury's elliptical orbit. (a) Determine the range of the Sun's gravitational field strength over the course of a Mercurian year. Use MSun = 1.989 1030 kg. gmin = Incorrect: Your answer is incorrect. N/kg gmax = Incorrect: Your answer is incorrect. N/kg (b) Write a symbolic expression for the ratio of the maximum and minimum gravitational field strengths that depends only on the distances, dmax and dmin, then calculate its numeric value. gmax gmin = Incorrect: Your answer is incorrect. = Incorrect: Your answer is incorrect. (c) The magnitude of the Sun's gravitational pull on Mercury is 8.989 1021 N when Mercury is at its farthest point. What magnitude gravitational force does the Sun exert on Mercury at its point of closest approach?
Explanation / Answer
(a)
Gravitational field strength is maximum when d is minimum
Using formula g = GM/r2
=> gmax = 6.67408 × 10-11 * 1.989 * 1030 / (4.600 * 1010)2 = 6.27351 * 10-2 N/Kg
Similarly, Gravitational field strength is minimum when d in maximum
=> gmin = 6.67408 × 10-11 * 1.989 * 1030 / (6.982 * 1010)2 = 2.72312 * 10-2 N/Kg
Range of Gravitational field strength over the course of a year = [2.72312 * 10-2, 6.27351 * 10-2]
(b)
From the formula gmax = GM/rmin2 and gmin = GM/rmax2
Dividing gmax by gmin, we get
=> gmax / gmin = (dmax / dmin)2
=> gmax / gmin = (6.982 * 1010 / 4.600 * 1010)2
=> gmax / gmin = 2.3038
(c)
At closest point, g is maximum
From the equation we derived in b
m * gmax / m * gmin = 2.3038
Fmax / Fmin = 2.3038
Given, Fmin = 8.989 * 1021 N
=> Fmax = 8.989 * 1021 * 2.3038 = 20.709 * 1021 N or 2.0709 * 1022 N
Gravitational force at the closest point = 20.709 * 1021 N or 2.0709 * 1022 N
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