Physics Problem set with Solution
Some of the posts in this blog include sample physics problem sets along with the post content. Readers can always try those. We have decided to publish the solutions to selected problems from different sets/posts, as we have done in this post.
Physics Problem Q set source: (based on the variation of g due to height or depth variation with respect to the earth’s surface)
If you want to go through the relevant tutorial and see the derivation of the formula used here then you can visit the link below.
Physics Problem & Solution -1
Q 1: What is the value of acceleration due to gravity at a height 4 miles above the earth’s surface?
data given: Diameter of the earth: 8000 miles.
g at the surface of the earth: 9.8 m/s^2.
Radius of the earth = 8000/2 = 4000 miles.
g = 9.8 m/s^2
The equation for acceleration due to gravity at a height 4 miles above is:
g1 = g (1 – 2h/R) = 9.8 [1 – (2X4/4000)] = 9.8 [1 – 2/1000] = 9.8 [ 1 – 0.002] = 9.78 m/s^2
Ans: 9.78 m/s^2
Physics Problem & Solution -2
Q2: At what depth under the earth’s surface, the value of acceleration due to gravity will reduce by 1% with respect to that on the earth’s surface?
Say, acceleration due to gravity on the earth surface is g and that at a depth h is g2.
g2 = g (1 – h/R) = g – gh/R
=> g – g2 =gh/R
=> Reduction of acceleration due to gravity = g -g2 = gh/R
According to the question this reduction, gh/R = 1 % of g
=> gh/R = (1/100) g = 0.01 g
=> h/R = 0.01
h = 0.01 R = 0.01 X 6400 Km = 64 Km
Answer: 64 km
Physics Problem & Solution -3
Q3: If the value of acceleration due to gravity at a small height h from the surface equals the value of acceleration due to gravity at a depth d, then find out the relationship between h and d.
At a height h,
g1 = g (1 – 2h/R) …………… (1)
At a depth d,
g2 = g (1 – d/R) ……………….(2)
according to the question, g1 = g2
=> g(1 – 2h/R) = g(1 – d/R)
=> 1 – 2h/R = 1 – d/R
2h = d
Answer: the relationship between h and d is 2h=d
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