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Gravitational potential at a point is equal to work done in bringing a unit mass from infinity to a particular point

That was the text book definition

$$V_{p} = -\frac{GM}{r}$$

If we calculate $V_p$ for the Earth surface with $R_e = 6.4 \times 10^6 \text{ m}$ and $M_e = 6 \times 10^{24} \text{ kg}$ we get

$$V_p = -6.35 \times 10^7 \text{ J/kg}$$

Q1) What does this number tell us? Does it mean I have to give a body of mass $1 \text{ kg}$ kinetic energy worth of $6.35 \times 10^7 \text{ J}$ in order for the body to leave this planet (escape Earth's gravitational influence) and after escaping Earth's influence its kinetic energy will be zero?

What does $-ve$ sign indicate?

Q2) Does higher magnitude of $V_{p}$ indicate higher potential energy of the body? If yes, we know that everything in this universe tries to reduce its potential so won't it make more sense if gravity doesn't attract but rather repel other mass as potential is zero at infinity, but it attracts? If higher magnitude of $V_{p}$ doesn't indicate higher potential energy of the body skip to next question.

Q3) We have two formulas to look at: $U = mgh$ and $V_p = -\frac{GM}{r}$. Let's deal with mass of $1 \text{ kg}$. Assuming that $-ve$ sign of gravitational potential has nothing to do with potential energy.

In the first formula we are multiplying by height, which means higher the body from the Earth surface higher is the potential which makes sense. Here $U$ and $h$ are directly proportional. The second formula doesn't make sense as we're diving with height, $V_p = -\frac{G M_e}{R_e+h}$ here $h$ and potential are inversely related?

I'm confused regarding this, tell me where am I going wrong

Gravitational potential at a point is equal to work done in bringing a unit mass from infinity to a particular point

That was the text book definition

$$V_{p} = -\frac{GM}{r}$$

If we calculate $V_p$ for the Earth surface with $R_e = 6.4 \times 10^6 \text{ m}$ and $M_e = 6 \times 10^{24} \text{ kg}$ we get

$$V_p = -6.35 \times 10^7 \text{ J/kg}$$

Q1) What does this number tell us? Does it mean I have to give a body of mass $1 \text{ kg}$ kinetic energy worth of $6.35 \times 10^7 \text{ J}$ in order for the body to leave this planet (escape Earth's gravitational influence) and after escaping Earth's influence its kinetic energy will be zero?

What does $-ve$ sign indicate?

Q2) Does higher magnitude of $V_{p}$ indicate higher potential energy of the body? If yes, we know that everything in this universe tries to reduce its potential so won't it make more sense if gravity doesn't attract but rather repel other mass as potential is zero at infinity, but it attracts? If higher magnitude of $V_{p}$ doesn't indicate higher potential energy of the body skip to next question.

Q3) We have two formulas to look at: $U = mgh$ and $V_p = -\frac{GM}{r}$. Let's deal with mass of $1 \text{ kg}$. Assuming that $-ve$ sign of gravitational potential has nothing to do with potential energy.

In the first formula we are multiplying by height, which means higher the body from the Earth surface higher is the potential which makes sense. Here $U$ and $h$ are directly proportional. The second formula doesn't make sense as we're dividing with height, $V_p = -\frac{G M_e}{R_e+h}$ here $h$ and potential are inversely related?

I'm confused regarding this, tell me where am I going wrong

Gravitational potential at a point is equal to work done in bringing a unit mass from infinity to a particular point 

That was the text book definition  $$V_{p} = -\frac{GM}{r}$$ if we try to calculate $V_{p} $ for earth surface
$(radius)_{earth} = 6.4 * 10^6 meters$
$(mass)_{earth} = 6*10^{24}kg$
now if we calculate $V_{p}$ $$V_{p} = -6.35 * 10^7\frac{J}{kg}$$

now what does this number tell us? does it mean I have to give a body of mass 1kg, Kinetic energy worth of $6.35*10^7$J in order for the body to leave this planet/escape earths gravitational influence and after escaping earths influence its KE will be zero?
What does $-ve$ sign indicate?
Q2)does higher magnitude of $V_{p}$ indicate higher potential energy of the body? if yes, we know that everything in this universe tries to reduce its potential so won't it make more sense if gravity doesn't attract but rather repels other mass as potential is zero at infinity, but it attacts. If higher magnitude of $V_{p}$ doesn't indicate higher potential energy of the body skip to next question

Q3
we have two formulas to look at
1. $U = mgh$
2. $V_{p} = -\frac{GM}{r}$
lets deal with mass of 1kg
assuming that -ve sign of gravitational potential has nothing with do with potential energy
In the first formula we're multiplying by height, which means higher the body from earth surface higher is the potential which makes sense. Here $U$ and $h$ are directly proportional
The second formula doesn't make sense as we're diving with height, $V_{p} = -\frac{GM_{earth}}{R_{earth}+h}$ here $h$ and potential are inversely related?
I'm confused regarding this, tell me where am I going wrong

Gravitational potential at a point is equal to work done in bringing a unit mass from infinity to a particular point

That was the text book definition 

$$V_{p} = -\frac{GM}{r}$$

If we calculate $V_p$ for the Earth surface with $R_e = 6.4 \times 10^6 \text{ m}$ and $M_e = 6 \times 10^{24} \text{ kg}$ we get

$$V_p = -6.35 \times 10^7 \text{ J/kg}$$

Q1) What does this number tell us? Does it mean I have to give a body of mass $1 \text{ kg}$ kinetic energy worth of $6.35 \times 10^7 \text{ J}$ in order for the body to leave this planet (escape Earth's gravitational influence) and after escaping Earth's influence its kinetic energy will be zero?

What does $-ve$ sign indicate?

Q2) Does higher magnitude of $V_{p}$ indicate higher potential energy of the body? If yes, we know that everything in this universe tries to reduce its potential so won't it make more sense if gravity doesn't attract but rather repel other mass as potential is zero at infinity, but it attracts? If higher magnitude of $V_{p}$ doesn't indicate higher potential energy of the body skip to next question.

Q3) We have two formulas to look at: $U = mgh$ and $V_p = -\frac{GM}{r}$. Let's deal with mass of $1 \text{ kg}$. Assuming that $-ve$ sign of gravitational potential has nothing to do with potential energy.

In the first formula we are multiplying by height, which means higher the body from the Earth surface higher is the potential which makes sense. Here $U$ and $h$ are directly proportional. The second formula doesn't make sense as we're diving with height, $V_p = -\frac{G M_e}{R_e+h}$ here $h$ and potential are inversely related?

I'm confused regarding this, tell me where am I going wrong