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I would like to know if an object will fall towards the sun if its horizontal speed is $0$ at earth's distance.

$G =$ gravitational constant
$m_1 =$ mass of object = 70 kg
$m_2 =$ mass of sun $= 1.989 \cdot 10^{30}$ kg
$r =$ Distance of object from sun = distance of earth to sun $= 149.6\cdot 10^6$ km

Assume there is no 'horizontal speed' (the velocity of $m_1$ tangent to $r$) and there is nothing else in the solar system.

Does this mean that by using

$$F = \frac{Gm_1m_2}{r^2}$$

There will be a Force of $0.4$ N towards the sun applied to the object. Which means it will start to 'fall' towards the sun with acceleration of $0.0057 $ m/s$^2$ (using $F=ma$)

In reality, does this mean a stationary (relative to the sun) astronaut at earth's distance will get pulled into the sun slowly?

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  • $\begingroup$ If there is no component tangent to the orbit, the body will definitely fall towards the sun. $\endgroup$
    – Sam
    Jan 15, 2020 at 6:28

4 Answers 4

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Yes, indeed it does.

The astronaut, in this scenario, is above a fixed (I'm going to ignore the Sun's rotation for simplicity) point on the Sun's surface, even if the "height" is profound (150 gigameters) and so begins to fall. For the same reason that if you are placed at a fixed point of height about the Earth's surface, and are released, you would begin to fall toward it. Gravity works the same way at any distance, it's just that it gets ever weaker.

The time required to fall can be found by Kepler's third law, because we can consider the fall to be an extreme case of an elliptical orbit (degenerated to a bicovered line segment, with eccentricity exactly 1). That is,

$$T_\mathrm{fall} = \frac{1}{2} \sqrt{\left(\frac{\pi^2}{2GM}\right) r^3}$$

(taking the semi-major axis as half the distance $r$ from the Sun, then half again because only half the "orbit" is completed) which gives a fall time of around 5.6 megaseconds (Ms), or roughly 1/6 of a year (~32 Ms). Hence your astronaut will, without special provision, die from running out of oxygen, or with only some provision, out of water, or with more but still not enough, out of food, in that order, before hir corpse is cremated in the Sun.

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  • $\begingroup$ Minor point. You've calculated the full orbital period for a 0.5 au semimajor axis. Fall time into the sun can be roughly estimated as half that, or about a sixth of a year. $\endgroup$
    – notovny
    Jan 15, 2020 at 11:09
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    $\begingroup$ @notovny: Thanks, yes I forgot about the missing "return trip" :) Fixed. $\endgroup$ Jan 15, 2020 at 11:11
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Yes, when something goes into orbit it needs to gain a lot of speed, if it doesn't have enough already.

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You should check your math. Convert all units to kg,m,and sec. Your basic understanding is correct. Unless there is some velocity perpendicular to the line from the object to the sun, the object will fall straight toward the sun. Although the acceleration is relatively small, the speed will build up pretty quickly.

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Without a perpendicular velocity, an object should fall directly towards the sun, because the only force pulls an object towards the center of the sun.

This means that at the distance Earth is from the Sun, an astronaut would also experience acceleration towards the Sun.

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