#Would the Earth suddenly start spinning differently? Yes, the barycenter (center of mass) of the earth-object system would move to a point between them. They would then spin around this point. The greater the altitude of the object with respect to the surface of the Earth, the greater the effect. The barycenter of a 2 object system is defined as:

##If the object were at an altitude of 0 meters, the orbital period of The Object and Earth would be 21 minutes. At an altitude of 35,768 km (Geosynchronous Orbit), the period would be 15 hours. How's that for a shortened day? Note: I ignored the gravitational effects of such an object this close to The Earth since OP specified:

#Would it start orbiting away from the Sun?

##No, it'll drop MUCH closer to the Sun.

###Therefore, after the object appears, the combined Earth-Object System would lose half its velocity under conservation of momentum.

##Therefore Earth would swing into orbit 5 million kilometers above Mercury. Needless to say, this would be BAD. World on fire BAD.

Would the Earth suddenly start spinning differently?

Yes, the barycenter (center of mass) of the earth-object system would move to a point between them. They would then spin around this point. The greater the altitude of the object with respect to the surface of the Earth, the greater the effect. The barycenter of a 2 object system is defined as:

If the object were at an altitude of 0 meters, the orbital period of The Object and Earth would be 21 minutes. At an altitude of 35,768 km (Geosynchronous Orbit), the period would be 15 hours. How's that for a shortened day?

Note: I ignored the gravitational effects of such an object this close to The Earth since OP specified:

Would it start orbiting away from the Sun?

No, it'll drop MUCH closer to the Sun.

Therefore, after the object appears, the combined Earth-Object System would lose half its velocity under conservation of momentum.

Therefore Earth would swing into orbit 5 million kilometers above Mercury. Needless to say, this would be BAD. World on fire BAD.

$r_1 = \cfrac{a}{1 + \cfrac{M_1}{M_2}}$

where $r_1$ is the distance from $M_1$ to the center of the system, $a$ is the distance between the two objects, and $M$ is the respective mass of each object. If in our case we assume their masses are identical and that $M_1$ is the earth, then: $r_1 = \cfrac{r_e + h}{1 + \cfrac{M_e}{M_e}} = \cfrac{1}{2}\left(r_e + h\right)$.

Furthermore, given Kepler's Laws of Motion, the period ('day' length) of this orbit would be: $T = \sqrt{\cfrac{4 \pi^2 a^3}{G\left(M_1 + M_2\right)}} = \sqrt{\cfrac{4 \pi^2 \left(r_1 + h\right)^3}{2GM_e}}$

##If the object were at an altitude of 0 meters, the orbital period of The Object and Earth would be 21 minutes. At an altitude of 35,768km (Geosynchronous Orbit), the period would be 15 hours. How's that for a shortened day? Note: I ignored the gravitational effects of such an object this close to The Earth since OP specified:

$F_G = 2ma = G\cfrac{2m M}{r^2} \Rightarrow a = G\cfrac{M}{r^2}$

which is the normal acceleration of the earth.

$m_1i v_1i = m_1f v_1f + m_2f v_2f$; m_1 = m_2 = m_e; v_1f = v_2f ; v_1i = v_e

###Therefore, after the object appears, the combined Earth-Object System would lose half it's velocity under conservation of momentum.

$$r_1 = \cfrac{a}{1 + \cfrac{M_1}{M_2}}$$

where $r_1$ is the distance from $M_1$ to the center of the system, $a$ is the distance between the two objects, and $M$ is the respective mass of each object. If in our case we assume their masses are identical and that $M_1$ is the earth, then:

$$r_1 = \cfrac{r_e + h}{1 + \cfrac{M_e}{M_e}} = \cfrac{1}{2}\left(r_e + h\right)$$.

Furthermore, given Kepler's Laws of Motion, the period ('day' length) of this orbit would be: $$T = \sqrt{\cfrac{4 \pi^2 a^3}{G\left(M_1 + M_2\right)}} = \sqrt{\cfrac{4 \pi^2 \left(r_1 + h\right)^3}{2GM_e}}$$

##If the object were at an altitude of 0 meters, the orbital period of The Object and Earth would be 21 minutes. At an altitude of 35,768 km (Geosynchronous Orbit), the period would be 15 hours. How's that for a shortened day? Note: I ignored the gravitational effects of such an object this close to The Earth since OP specified:

$$F_G = 2ma = G\cfrac{2m M}{r^2} \Rightarrow a = G\cfrac{M}{r^2}$$

which is the normal acceleration of the Earth.

$m_1i v_1i = m_1f v_1f + m_2f v_2f; m_1 = m_2 = m_e; v_1f = v_2f ; v_1i = v_e$

###Therefore, after the object appears, the combined Earth-Object System would lose half its velocity under conservation of momentum.

If the height is higher, than the barycenter would move into the mantle. Given Kepler's Laws of Motion  , the period of the orbit ('day' length) would be: $T = \sqrt{\cfrac{4 \pi^2 a^3}{G\left(M_1 + M_2\right)}} = \sqrt{\cfrac{4 \pi^2 \left(r_1 + h\right)^3}{2GM_e}}$

##Given a height of 0 meters, the orbital period of The Object and Earth would be 59.64 min. That's a 1 hour day.

But this answer is not quite as simple. The mass of the 'Earth' system will has doubled, therefore, he force of gravity upon the system from the sun would also double:

$F_{G_2} = G\cfrac{2M_1 M_2}{r^2} = 2F_{G_1}$

  where

 $F_{G_1} = G\cfrac{M_1 M_2}{r^2}$.

but, since newton's 2nd law states:

$F = ma$

  then the combination would yield:

which is the normal acceleration on earth.

Instead, the sudden increase in mass on earth ($m \rightarrow 2m$) would create a change in momentum (impulse); however, momentum must remain conserved since no outside force acted upon the system.

$m_i v_i = m_f v_f$ and $m_f = 2m_i$

$m_i v_i = 2m_i v_f \Rightarrow v_i = 2v_f \Rightarrow v_f = \frac{1}{2} v_i$

Using the Vis-Viva equation, the new semimajor axis of the elliptical orbit would become:

plugging in some average numbers ($v = 30 \space\text{km/s}$, $r = 150 \times 10^6 \space\text{km}$, $M = 2 \times 10^{30} \space\text{kg}$) yields $7.5 \times 10^7 \space\text{km}$ or $75 \times 10^6 \space\text{km}$. For comparison, Venus orbits around $108 \times 10^6 \space\text{km}$ from the sun, and Mercury orbits between $46\times 10^6 \space\text{km}$ and $70 \times 10^6 \space\text{km}$ from the Sun.

##Therefore Earth would swing into orbit 5 million kilometers above Mercury. Needless to say, this would be BAD. World on fire, atmosphere boiling BAD.

But hey! We'd have a few months to kill before The End!

If the height is higher, than the barycenter would move into the mantle.

Furthermore, given Kepler's Laws of Motion, the period ('day' length) of this orbit would be: $T = \sqrt{\cfrac{4 \pi^2 a^3}{G\left(M_1 + M_2\right)}} = \sqrt{\cfrac{4 \pi^2 \left(r_1 + h\right)^3}{2GM_e}}$

##If the object were at an altitude of 0 meters, the orbital period of The Object and Earth would be 21 minutes. At an altitude of 35,768km (Geosynchronous Orbit), the period would be 15 hours. How's that for a shortened day? Note: I ignored the gravitational effects of such an object this close to The Earth since OP specified:

and for some reason, doesn't break it.

This answer is not quite as simple due to orbital mechanics. The mass of the 'Earth' system will has doubled, therefore, the force of gravity upon the system from the sun would also double:

$F_{G_2} = G\cfrac{2M_1 M_2}{r^2} = 2F_{G_1}$ where  $F_{G_1} = G\cfrac{M_1 M_2}{r^2}$;

but, since newton's 2nd law states  $F = ma$ then the combination yields:

which is the normal acceleration of the earth.

Instead, the sudden increase in mass on earth ($m \rightarrow 2m$) would create a change in momentum (impulse). Now, several assumptions must be made, all of whom are heavily dependent on the object's origin. I assume the following:

1. The Object 'appeared' out of nowhere.
2. The Object and The Earth may be considered a single system with respect to The Sun.

Given conservation of momentum:

$m_1i v_1i = m_1f v_1f + m_2f v_2f$; m_1 = m_2 = m_e; v_1f = v_2f ; v_1i = v_e

$m_e v_e = 2m_e v_f \Rightarrow v_e = 2v_f \Rightarrow v_f = \frac{1}{2} v_e$

###Therefore, after the object appears, the combined Earth-Object System would lose half it's velocity under conservation of momentum.

Using the Vis-Viva equation, we may then calculate what the semimajor axis of the elliptical orbit would become (it won't but bear with me here):

Great, now we can figure out the semimajor axis of the new orbit. Since Earth has an elliptical orbit, our orbital velocity, $v_e$ varies over the year. At aphelion (when we're farthest from The Sun in July), $v_e = 29.29 \space\text{km/s}$. At perihelion (when we're closest to The Sun in January) $v_e = 30.29 \space\text{km/s}$ (source: NASA). Since they have a 3.3% difference, we'll simplify our calculations and just go with $v = 30 \space\text{km/s}$. We'll also take $r = 150 \times 10^6 \space\text{km}$, $M = 2 \times 10^{30} \space\text{kg}$).

This yields $75 \times 10^6 \space\text{km}$ or 0.5 AU. For comparison, Venus orbits around $108 \times 10^6 \space\text{km}$ from the sun, and Mercury orbits between $46\times 10^6 \space\text{km}$ and $70 \times 10^6 \space\text{km}$ from the Sun. Since this would be the new stable orbit, the Earth-Object system would naturally tend towards this orbit; however, it's still at 1 AU.

##Therefore Earth would swing into orbit 5 million kilometers above Mercury. Needless to say, this would be BAD. World on fire BAD.

This is BAD for a multitude of reasons:

1. At this distance, all water on Earth would boil. The oceans, the seas, etc.
2. UV radiation woulld quadruple.
3. While swinging into this orbit, the orbits of Mecury and Venus would also be altered. If The Earth and Mercury were to get close enough, a collision would be a significant possibility.
4. Since Earth is swinging into this orbit (0.5 AU) from its original orbit (1AU), it'll swing back out of it. Then it'll swing back in, and repeat until the orbit has stabilized.

Note: I used Kepler's laws and convservation of momentum to make simplifying assumptions. The reality is that I'd need to simulate this to come up with an accurate representation.