# Solving Gravitational Field: What Did I Do Wrong?

• UrbanXrisis
In summary, the conversation discusses the calculation of gravitational field g for a system with two masses, represented by the equation g=\frac{2MGr}{(r^2+a^2)^{3/2}}. The question arises as to why the book's answer differs from the calculated value of g=2\frac{GM}{r^2}. It is determined that this calculation is incorrect and the correct value is g=\frac{2GM\cos{\alpha}}{a^2 + r^2}, where \cos{\alpha}=\frac{r}{a^2+r^2}.
UrbanXrisis

$$g=\frac{GM}{x^2}$$
$$x=\sqrt{r^2+a^2}$$
$$g=\frac{GM}{(\sqrt{r^2+a^2})^2}$$
$$g=\frac{GM}{r^2+a^2}$$

since there are 2 masses...
$$g=2 \frac{GM}{r^2+a^2}$$

$$g=\frac{2MGr}{(r^2+a^2)^{3/2}}$$

what did I do wrong?

Last edited by a moderator:
UrbanXrisis said:
$$g=\frac{GM}{r^2+a^2}$$
OK, this is the field due to one of the masses. Note that it's a vector. What is its direction?

since there are 2 masses...
$$g=2 \frac{GM}{r^2+a^2}$$
Since the field contributions are vectors, they must be added as such. (You can double it only if the vectors were in the same direction.) Hint: The vertical components will cancel.

horizontally:
$$g=\frac{GM}{r^2}+\frac{GM}{r^2}$$
vertically:
$$g=\frac{GM}{a^2}-\frac{GM}{a^2}=0$$

so then g would be: $$g=2\frac{GM}{r^2}$$

not quite sure what to do

That's WRONG.
Vertically:
$$\frac{GM\sin{\alpha}}{a^2 + r^2} - \frac{GM\sin{\alpha}}{a^2 + r^2} = 0$$
Horrizontally:
$$\frac{GM\cos{\alpha}}{a^2 + r^2} + \frac{GM\cos{\alpha}}{a^2 + r^2} = \frac{2GM\cos{\alpha}}{a^2 + r^2}$$
Find $\cos{\alpha}$. Can you handle it from here?
Viet Dao,

$$\cos{\alpha}=\frac{r}{a^2+r^2}$$

$$g=\frac{2GM\frac{r}{a^2+r^2}}{a^2 + r^2}$$

?

That's not correct. Recheck your $cos{\alpha}$.
Viet Dao,

## 1. What is a gravitational field?

A gravitational field is a region of space around an object where other objects with mass experience a force of attraction towards it. The strength of this force depends on the mass of the object creating the gravitational field.

## 2. How do I calculate the strength of a gravitational field?

The strength of a gravitational field can be calculated using the formula F = G(m1m2)/r^2, where F is the force of attraction, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. This formula is known as Newton's law of universal gravitation.

## 3. What are common mistakes when solving for gravitational field?

Some common mistakes when solving for gravitational field include using the incorrect formula, using the wrong units, and not properly considering the direction of the force. It is important to carefully read the problem and double check all calculations to avoid these errors.

## 4. What are the units for gravitational field?

The units for gravitational field are Newtons per kilogram (N/kg) in the SI system. In the imperial system, the units are pounds per slug (lb/slug). These units represent the force per unit mass experienced by an object in a gravitational field.

## 5. How does distance affect gravitational field?

According to the formula F = G(m1m2)/r^2, the strength of a gravitational field is inversely proportional to the square of the distance between the objects. This means that as the distance between two objects increases, the force of attraction decreases. Therefore, the farther an object is from a massive body, the weaker the gravitational field will be.

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